Circle Inscribed in a Parabola

Maximum area of circle holds when the circle tangent the parabola, which mean $D = 0$ . So, Since the circle centered at the origin, then the equation of circle is $x^2 + y^2 = r^2$ . So, from $y = x^2 - 100$ , we obtain $x^2 = y + 100$ . Subtitute into the circle equation :

$x^2 + y^2 = r^2$ $y + 100 + y^2 = r^2$ $y^2 + y + 100 - r^2 = 0$ , since the circle tangent the parabola, $D = 0$ , implies $b^2 - 4ac = 0 \implies 1 - 4(100 - r^2) = 0 \implies r^2 = \frac{399}{4}$

Then, area of circle $= \pi. r^2 = \pi. \frac{399}{4}$

So, $a + b = 399 + 4 = \boxed{403}$

Shortest line from origin to a point on the curve must be normal to the curve. $\begin{aligned} & -\frac{1}{\frac{\partial \left( {{x}^{2}}-100 \right)}{\partial x}}=\frac{{{x}^{2}}-100}{x} \\ & -\frac{1}{2x}=\frac{{{x}^{2}}-100}{x} \\ & x=\pm \sqrt{\frac{199}{2}} \end{aligned}$

Hence shortest line is $\begin{aligned} & \sqrt{{{x}^{2}}+{{y}^{2}}} \\ & =\sqrt{{{\sqrt{\frac{199}{2}}}^{2}}+{{\left( {{\sqrt{\frac{199}{2}}}^{2}}-100 \right)}^{2}}} \\ & =\frac{\sqrt{399}}{2} \\ \end{aligned}$ Area of circle $\begin{aligned} & \pi {{r}^{2}} \\ & =\pi {{\left( \frac{\sqrt{399}}{2} \right)}^{2}} \\ & =\frac{399\pi }{4} \\ \end{aligned}$

If the circle is inscribed in the parabola, then they share a common tangent.

(1) x^2 + y^2 = r^2

(2) y = x^2 - 100

Both of these equations must have the same slope at the points of intersection (since they both have a common tangent)

dy/dx for (1) is -x/y

dy/dx for (2) is 2x

-x/y = 2x, which leads to y = -1/2 y = x^2 - 100 --> x^2 = 99 1/2 = 398/4

x^2 + y^2 = 399/4 = r^2, which leads to Area = pi * r^2 = 399/4 pi

399 + 4 = 403

Since the circle is inscribed, it hits the parabola at two points, and the x-coordinates of these points are additive inverses. Let the radius of the circle be $r$ , and so the answer we are looking for is $\pi r^2$ . We must find $r$ when this system has two solutions: $y=x^2-100$ $x^2+y^2=r^2$ Solving for $y$ in both, we get: $y=x^2-100=\pm\sqrt{r^2-x^2}$ Now, we have eliminated $y$ , and can simplify: $x^2-100=\pm\sqrt{r^2-x^2}$ $x^4-200x^2+10000=r^2-x^2$ $x^4-199x^2+10000-r^2=0$ We can treat this as a quadratic in $x^2$ , and each solution for $x^2$ gives two solutions for $x$ . If we want two solutions for $x$ , there must be one solution for $x^2$ , and so the discriminant must be 0. $199^2-4(1)(10000-r^2)=0$ $39601-40000+4r^2=0$ $4r^2=399$ $r^2=\frac{399}{4}$ Therefore, the area of the circle is $\frac{399}{4}\pi$ , and the answer is $399+4=403$ .

Because the circle centered at the origin, which is inscribed in parabola, so we have the point centre of the circle (0;0) and equation is x^2+y^2= r^2.

after that, the circle and the parabola must have the same slopes in 2 points of contact.

x^2+y^2=r^2 => dy/dx= -x/y and y=x^2-100 => dy/dx= 2x

From 2 these equations we obtain y = -1/2 and x^2 = 100-1/2

x^2+y^2= 100-1/2+1/4=399/4

finally, a/b=399/4 so a+b=403

As both the parabola and the circle are symmetric about the y-axis , the circle will touch the parabola in two points and the y-coordinate of both these points will be same . Let ( h , k ) be the point at which the circle touches the parabola . Here , h will assume two values whereas k will assume only one . Since this point lies both on the parabola and the circle , it will satisfy the parabola's as well as the circle's equation . Equation of the circle centred on the origin is x ^2 + y ^2 = r ^2 , where r is the radius of the inscribed circle . Therefore , h ^2 + k ^2 = r ^2

\Rightarrow h ^2 = r ^2 - k ^2 ......eq-1

Also, k = h ^2 - 100 .......eq-2

Substituting eq-1 in eq-2 , we get

k = r ^2 - k ^2 -100

\Rightarrow k ^2 + k +100 - r ^2 = 0 , which is a quadratic in k . To get one real and distinct value of k , we should have one root of the above obtained quadratic , which is only possible if the discriminant D = 0 . Hence , 1^2 - 4(1)(100 - r ^2) = 0

\Rightarrow 1 - 400 + 4 r ^2 = 0

\Rightarrow 4 r ^2 = 399

\Rightarrow r ^2 =\frac{399}{4}

Thus , the area of the area of the circle is \pi * r ^2 = \frac{399}{4}\pi Therefore , a + b = 399 + 4 = 403

As the circle is inscribed in the parabola so both of them will have common tangents at the point of contact. Equation of tangent in point form of circle and parabola are respectively:

       xh+ky=R^{2}
       xh-y/2=k/2+100

where R is the radius of circle and (h,k) are the points contact of parabola and circle and also the point at which tangent is drawn.

As the both are the same, so on comparing the coefficient of variable(x & y) we get:

k=-1/2

Now comparing the constant terms on RHS we get:

R^{2}=399/4

Since, area of circle = π.R^{2} = π.399/4

Hence, a+b=403

The circle leads to the equation: $x^2+y^2=r^2$ (1) The circle and the parabola have 2 common tangent points, which are reflected by each other through the line Oy, hence $y 1=y 2$ and $x 1=-x 2$ Since the tangent points are on the parabola, then $y=x^2-100$ Substitute this in (1) gives $y^2+y+100-r^2=0$ This equation must have one and only root since $y 1=y 2$, then $1-4(100-r^2)=0$, thus $\frac{a}{b}=r^2=\frac{399}{4}$ The answer is 403.

Imgur Consider perpendicular TM dropped on the axis from the point of tangency T. It can be shown that the distance from the center of the circle O to M is always 1/2 (isn't that fascinating?!) unless of course the radius is smaller than 1/2 (the radius of curvature at the vertex)

Thus OM = 1/2

Substitute y = -1/2 in the parabola's equation to get x coordinate of T which gives -

MT = sqrt(199/2)

Using Pythagoras OT ^ 2 = 399/4

This solution uses a bit of calculus.

In order to find the area of the circle, we need to figure out the radius. Since the circle is inscribed in the parabola, the radius is equal to the shortest distance from the origin to any point on the parabola.

Starting with the distance formula:

$r = \mathrm{min}\left(\sqrt{x^2 + y^2}\right)$

$r = \mathrm{min}\left(\sqrt{x^2 + (x^2 - 100)^2}\right)$

The minimum of the function will be at one of the functions critical points (i.e., where the derivative is zero or undefined).

$\frac{d}{dx} \sqrt{x^2 + (x^2 - 100)^2} = 0 \: \mathrm{or \: undefined}$

$\frac{1}{2}(x^2 + (x^2 - 100)^2)^{-1/2}(2x + 2(x -100)(2x) = 0 \: \mathrm{or \: undefined}$

$\frac{1}{2}(x^2 + (x^4 - 200x^2 + 10000)^{-1/2}(4x^3 - 398x) = 0 \: \mathrm{or \: undefined}$

$\frac{2x^3 - 199x}{\sqrt{x^4 - 199x^2 + 10000}} = 0 \: \mathrm{or \: undefined}$

$2x^3 - 199x = 0 \: \: \: \mathrm{or} \: \: \: \sqrt{x^4 - 199x^2 + 10000} = 0$

$x = \pm \sqrt{\frac{199}{2}}, 0$

(The right equation had no real solutions.)

We could also use calculus to check if these values are maxima or minima, but it's easier to just plug the values back into the original distance equation:

At $x=\sqrt{\frac{199}{2}}$ , distance is $\sqrt{\frac{399}{4}}$ .

At $x=-\sqrt{\frac{199}{2}}$ , distance is $\sqrt{\frac{399}{4}}$ .

At $x=0$ , distance is $100$ .

Therefore, the minimum distance (and therefore the radius) is $\sqrt{\frac{399}{4}}$ (at $x = \pm \sqrt{\frac{199}{2}}$ , and the area of the circle is $\frac{399}{4} \pi$ .

Finally, $399 + 4 = \fbox{403}$ .

The parabola is described by $y=x^2-100$ , while the circle is described by $x^2 + y^2 = r^2$ , since it's centered at the origin. It is easy to see that the circle is inscribed in the parabola iff they intersect each other in exactly 2 distinct points, which are symmetric respect to the y-axis. So in order to calculate r, we equalize the equations and get $y + y^2 = r^2 - 100$ or $y^2 + y + (100 - r^2) = 0$ We want this equation to have exactly one solution for y. That's the case if the discriminant is equal to 0. $1 + 4(r^2 - 100) = 0$ => $r = \sqrt{\frac{399}{4}}$ The area A of the circle is $r^2\pi$ , so we have $A = \frac{399}{4}\pi$ and finally $a + b = 403$

Clearly,the vertex of the parabola is (0,-100). So, the parabola is symmetrical about y-axis. Also, as the circle is centered at origin, it must be symmetrical about y-axis. The circle will be inscribed inside parabola if it touches the parabola. As it is opening upward parabola, circle centered at origin will touch the parabola at exactly 2 points. By symmetry we can see that the y co-ordinate of both the points must be same.Now, let the radius of circle be r. The equation of circle becomes $x^2 + y^2 = r^2$ . Substituting $x^2$ from the given equation of parabola we get $y^2 + y +100 - r^2 = 0$ . Now using the method of completing the square, we add and subtract 1/4 to the equation and it becomes, $(y + 1/2)^2 + 399/4 - r^2 = 0$ . As y can have only one value $399/4 - r^2$ must be $0$ . So we get $a/b = 399/4$ and $a+b = 403$ is the required answer.

NOTE: I am editing muhammad's solution

Circle is inscribed in the parabola so it touches the parabola at two points of the form (a,b) and (-a,b). { Note that these points may not be distinct i.e.(0,0) }

Let r be the radius of the inscribed circle.

Now , Since the circle is centered at the origin, then the equation of circle is x^2+y^2=r^2. So, from y=x^2−100, we obtain x^2=y+100. Substitute into the circle equation :

x^2+y^2=r^2

y+100+y^2=r^2

y^2+y+100−r^2=0

roots of this equation gives the value of y coordinates of two given points {i.e (-a,b) and (a,b) }

since roots are equal, D=0, implies

b^2−4 a c = 0 = 1−4(100−r^2)=0⟹r^2=399/4

Then, area of circle = π .r^2 = π * 399/4

So, a+b=399+4=403

By the equation of the circle we have: x^2 +y^2= r^2 (1) , where r is the radius. But for the two points of the parabola: y= x^2 - 100 (2). With (2) in (1) we have: x^2 + (x-100)^2= r^2. Therefore , x^4 -199x^2 +10000 - r^2 = 0. But x must be a real number , and there are only two points of contact between the parabola and the circle , thus: (199)^2 -4(10000- r^2)=0 . This implies that : r^2 = 399/4 and the area is 399/4*pi , so a + b = 403.

Since the circle is inscribed inside the parabola, the radius of the circle is equal to the shortest distance between the center of the circle (i.e is origin) and the parabola. => distance between the center of the circle and parabola = (x^2 + y^2)^(1/2) = (x^2 + (x^2-100)^2 ) ^(1/2) = (x^4-199x^2+10000)^(1/2) =( (x^2)^2 -2*x^2 * 199/2 + (199/2)^2 - (199/2)^2 +10000 ) ^(1/2) = ( (x^2-199/2)^2 +399/4) )^(1/2) Now, (x^2 - 199/2)^2 >=0 for any real value of x , (x^2 -199/2)^2 +399/4 >= 399/4 , ( (x^2 -199/2)^2 +399/4) )^(1/2) >= (399/4)^(1/2). Thus, shortest distance between the origin and the parabola = (399/4)^1/2 => radius of the circle (r) = (399/4)^(1/2) AREA OF THE CIRCLE = pi r^2 = 399/4 pi => a+b = 399+4 = 403

Lagrange multiplier can be used to solve the problem. it is not necessarily a simple approach, but it could provide a different way to look at the problem.

If the circle intersects the parabola, then area or $r^{2}$ is smallest when the circle is inscribed in the parabola. So the problem can be treated as a constrained optimization problem. Then constraint is $g=x^{2}-y=100$ , and the function to minimize is $f=r^{2}=x^{2}+y^{2}$ .

The gradient of g is <2x, -1>, and the gradient of f is <2x, 2y>, the maximum happens when the two gradient is parallel to each other(equivalent to f and g is tangent to each other). Solving $f=\lambda g$ and the constraint $x^{2}-y=100$ , we have $y=-\frac{1}{2}$ . Substituting into the parabola we have $x^{2}=100-\frac{1}{2}$ . So $r^{2}=x^{2}+y^{2}=100-\frac{1}{2}+\frac{1}{4}=\frac{399}{4}$ . So $\frac{a}{b}=r^{2}=\frac{399}{4}$ , $a + b = 399 + 3=\boxed{403}$ .

We have distance from right-side points of the parabola to the origin is $d^2 = x^2 + y^2 = x^2 + (x^2 - 100)^2 = g(x)$ . So the circle touches parabola at point which has smallest $g(x)$ . Or $\frac{dg(x)} { dx } = 0 \Leftrightarrow x^2 = 100 - 1/2~and~y=-1/2$ . From that we have $a + b = \boxed{403}.$

As we know, the equation of a circle centered at the origin is given by: $$ x^2 + y^2 = R^2 $$ We are looking for the circle that have least intersections with the parabola, using the equation of the circle we have: $$ (y + 100) + y^2 = R^2 $$ Making the discriminant $\Delta = 0$ we have the wanted radius: $$ \Delta = 0 \Rightarrow 1 - 4\cdot(100 - R^2) = 0 $$ Hence: $$ R^2 = \frac{399}{4} $$ Knowing that the area of the circle is given by $\pi R^2$ we finally get the answer: $$ a + b = 399 + 4 = 403 $$

First, recall the equation of a circle in the plane: $x^2 + y^2 = R^2$ . Next, recall what it means for a circle to be inscribed: it is the largest possible circle that can be drawn interior to a planar figure. Hence, this circle must be wholly contained by the parabola $y=x^2-100$ .

Using this information, we can solve for the general intersection of the curves and impose an extra restriction: they only intersect for one value of $y$ . To convince yourself that this must be the case, consider all possible number of intersections of a circle and a parabola: 0, 2, and 4.

So we want to find a value of $R$ such that the following quadratic equation has exactly one solution $\begin{aligned} 100 + y + y^2 &= R^2 \\ (100-R^2) + y + y^2 &= 0\\ \implies y &= \dfrac{1 \pm \sqrt{1 + 4R^2 - 100}}{2} \end{aligned}$

In order for this equation to have exactly one solution, it must be the case that $R^2=\dfrac{399}{4}$ . Noting that the area of such a circle is $\pi R^2 = \pi \dfrac{399}{4}$ , and noting that this fraction is coprime, we conclude that our answer is $399 + 4 = 403$

Note: It is possible for a circle to intersect with a parabola in two places and not satisfy the definition of being inscribed. This occurs when it is so big that it encompasses the entire bottom of the parabola. To explicitly discount this situation, we should require that $R$ not exceed 100.

You can solve this problem with good idea or creativity on the manipulation of that function. We know, that the function of circle in the origin is x^{2} + y^{2} = r^{2} , assuming that r is the radius of the circle.

Substitue this function into the parabola function followed by:

y = x^{2} -100

y = r^{2} - y^{2} - 100 <= x^{2} = r^{2} - y^{2}

Then, make it as the quadratic function like this:

y^{2} + y + (100 - r^{2}) = 0

completing the square of the function above, we get (y + \frac {1}{2})^{2} - \frac {1}{4} + 100 - r^{2} = 0 . Next, you can imagine that, when we submit y=0 then it must get the radius of the circle ( the picture is nice to explain it :)) )

And, you will get the r^2 = \frac {399}{4}* , so the area of the circle is * phi.r^{2} = \frac {399phi}/4 , * *a = 399 * and * b = 4 , so * *a + b = **403 .

Cicrle: x^2 + y^2 = R^2 is inscribed in the parabola y= x^2 -100, means R^2= y^2 +y +100 min. R^2=(y+1/2)^2 +399/4 >= 399/4, so a/b = 399/4 and a+b= 403

First, we need to find out the radius of the circle to calculate the area. To find the radius, we need to know the point where the circle touches the curve. At that point, the gradient will be the same.

So, let say the equation of the circle : $x^{2}$ + $y^{2}$ = $r^{2}$

By differentiating the equation of the circle, we get dy/dx = gradient = * $-\frac{x}{y}$ *

By differentiating the equation of the curve, we get dy/dx = gradient = 2x

Since the gradient is the same, hence :

2x = $-\frac{x}{y}$

y = $-\frac{1}{2}$

Using the curve equation, we get that the x = $\sqrt{398}$ / 2

Then, using the circle equation, we get that the $r^{2}$ = $\frac{399}{4}$

The area of the circle is $π \times r^{2}\$ = $\frac{399}{4}$ π

Hence, a+b = 403

Let P (x,y) represent the point on the parabola closest to the origin. Let r represent the distance from the origin to point P.

r = \sqrt {x^2 + y^2}, sub in y = x^2 - 100

r = \sqrt {x^2 + (x^2 - 100)^2}

r = \sqrt {x^4 - 199x^2 + 10000} (1)

r^2 = {x^4 - 199x^2 + 10000}

Do implicit differentiation:

2r*(r') = 4x^3 - 398x, set r' = 0

4x^3 - 398x = 0

x(2x^2 - 199) = 0

x = 0 or x^2 = \frac {199}{2}

Sub both values into equation (1) and we get distances of: r = 100 and r = \sqrt {\frac {399}{4}}

The second value is smaller in value, which gives us the radius of the inscribed circle.

The area is A = \pir^2

A = \frac {399}{4} \pi

Thus a = 399, b = 4, and a+b = 403

A circle with radius $r$ centered at the origin has an equation $x^2+y^2=r^2$ . Assume the circle and parabola intersects at point $(a,b)$ . The line tangent to the circle at point $(a,b)$ is the same as the line tangent to the parabola at point $(a,b)$ . Thus, the slopes of the tangent lines are the same.

The slope of the line tangent to the parabola is given by $m_1=y'=(x^2-100)'=2x$ . At $x=a$ , $m_1=2a$ . For the slope of the tangent line to the circle, we employ implicit differentiation. $(x^2+y^2)'=(r^2)' ,2x+2yy'=0 ,m_2=y'=-\frac{x}{y}$ . At $x=a$ and $y=b$ , $m_2=-\frac{a}{b}$ . Since $m_1=m_2$ , $2a=-\frac{a}{b}$ and $b=-\frac{1}{2}$ . $b^2=\frac{1}{4}$ and $a^2=100-\frac{1}{2}$ so that $a^2=\frac{199}{2}$ . Hence, $r^2=a^2+b^2=\frac{199}{2}+\frac{1}{4}=\frac{399}{4}$ . The area of the circle is given by $πr^2=\frac{399}{4}π$ . Hence, $a+b=399+4=403$

First, we can manipulate the equation of parabola: [; y = x^{2}-100\Leftrightarrow x^{2}=y+100 ;]

The equation of a circle with center in C(0, 0) is given by: [; (x-0)^{2}+(y-0)^{2}=r^{2} ;]

As the circle is tangent with parabola, we can use the manipulated equation of parabola and use in the circle equation: [; \left (y+100 \right )+y^{2}=r^{2}\Leftrightarrow y^{2}+y+(100-r^{2})=0 ;]

This quadratic equation in [; y ;] admits only one answer since the parabola have a horizontal guideline. With this fact, we can equal the discriminant to zero: [; \Delta = 0\Leftrightarrow \frac{1^{2}-4\cdot 1\cdot (100-r^{2})}{2\cdot 1}=0 ;]

Isolating [; r^{2} ;] in the last equation: [; 1-400-4r^{2}=0\Leftrightarrow r^{2}=\frac{399}{4} ;]

The area of the circle with radius [; r ;] is given by: [; A=r^{2}\cdot\pi=\left (\frac{399}{4} \right )\cdot\pi ;]

Where [; a = 399 ;] and [; b = 4 ;]. Then, [; a + b = 399 + 4 = 403 ;].

Equation of circle: (x^2) + (y^2) = (r^2) ; Equation of parabola: (x^2) = (y + 100) so if we want to find the meet point of these 2 curves => substitute x^2 = y + 100 into equation of circle => (y+100) + (y^2) = (r^2) this equation has 2 roots, which denote the two different values of 'y' for which the circle and parabola intersect... but both parabola and circle are perfectly symmetrical about y-axis so, it is obvious that the 2 meeting points should have same y-coordinate This simpy means that the equation: y^2 + (y+100) - r^2 = 0 has repeated roots => Discriminant = 0 => r^2 = 399/4 Area of circle = pi (r^2) = pi (399/4) ===> (a+b) = 399+4 = 403

y = x² - 100 x² = y + 100

The circle equation can take this form:

x² + y² - r² = 0, where r is the radius. y + 100 + y² - r² = 0 y² + y + (100 - r²) = 0

By symmetry, the graphs touch at two points having the same y-coordinate, so the discriminant of the quadratic equation must be zero.

1 - 4(100 - r²) = 0 4r² - 399 = 0 r² = 399/4 a/b = 399/4

Since 399 and 4 are coprime, and any integer multiples of them would not be, a = 399, b = 4 a + b = 403

circle centered at the origin is x^2 + y^2 = R^2

x^2 = R^2 - y^2 ...... (I)

and parabaola y = x^2 - 100

substitution with (I)

y= R^2 - y^2 - 100

y^2 + y + 100 - R^2 = 0

because the cirle is tangents with the parabola, determinant of y = 0

1 - 4(100 - R^2) = 0

R^2 = 99,75

The area of a circle is πR^2 = a/b.π

so a/b = R^2 = 99,75 = 399/4

a + b = 399 + 4 = 403

y= x^2 -100 y^2= R^2 - x^2 (a/b= R^2) => y^2 + y + 100 - R^2 = 0 this equation must have only 1 value of y to sastify the problem so 1-4(100-R^2)=0 <=> R^2=399/4 => a+b = 403

y = x² - 100 x² = y + 100

The circle equation can take this form:

x² + y² - r² = 0, where r is the radius. y + 100 + y² - r² = 0 y² + y + (100 - r²) = 0

By symmetry, the graphs touch at two points having the same y-coordinate, so the discriminant of the quadratic equation must be zero.

1 - 4(100 - r²) = 0 4r² - 399 = 0 r² = 399/4 a/b = 399/4

Since 399 and 4 are coprime, and any integer multiples of them would not be, a = 399, b = 4 a + b = 403

Find the contact point between the circle and the parabola. If that point has coordinate (x,y) then radius vector (x,y) must be perpendicular to tangential vector (1,dy/dx) = (1, 2x) The condition above can be express by the formula: x+(x^2-100)*2x=0 The two solution : x=0 or x^2=199/2 For x=0 the radius of the inscribed circle is 100 For x^2 = 199/2 ( then y=-1/2) the radius of the inscribed circle is \sqrt{399/4} We choose the smaller value ( if we choose the larger one that circle will cut the parabola) so a/b= R^2=399/4. So a+b=304

To find area of the circle, we first find the radius r.

By Pythagoras' Theorem,

$r$ = $\sqrt{y^{2}+x^{2}}$ .

Given that $y$ = $x^{2}-100$ ,

We can substitute the first equation into the second to get:

$r$ = $y^{2}+y+100$ = $0.25(2y+1)^{2} + \frac{399}{4}$

As the radius r is the shortest distance from the origin to the parabola, we must minimise the value of r, which is at $r$ = $\sqrt{\frac{399}{4}}$

Area of circle is thus equals $pi \times r^{2}$ = $\frac{399}{4} \times pi$

Hence $a+b$ = $399+4=403$

The radius of this inscribed circle can be found by calculating the minimal distance from the origin to the given parabola, that is $\min \sqrt{x ^ 2 + \left(x ^ 2 - 100\right) ^ 2}$ rearranging, we have $\min \sqrt{\left(x ^ 2 - \frac{199}{2}\right) ^ 2+\frac{399}{4}}$ which is $\frac{ \sqrt{399}} {2}$ and occurs at $x = \pm \sqrt{\frac{199}{2}}$ .

The area is therefore $\left(\frac{\sqrt{399}}{2}\right) ^ 2 \pi = \frac{399}{4} \pi$

A circle centered at the origin of radius $r$ has equation $x^2+y^2 = r^2$ . To find the intersection of this circle with the parabola, we substitute $x^2 = y+100$ (since we wish to simplify things and not make them horrible quartics) to get $y^2 + y + 100 - r^2 = 0$ . Since both the circle and the parabola are symmetric about the $y$ -axis, each solution to this quadratic corresponds to two points of intersection. Therefore, we want it to have only one solution, which means the discriminant must vanish. This gives us $1-4(100-r^2) = 0$ or $r^2 = \frac{399}{4}$ . Since the area is $\pi r^2$ , we have $399+4 = 403$ .

Differentiating the equation of the circle, $x^2 + y^2 = r^2$ , and the equation of the parabola, $y = x^2 - 100$ , we obtain the equations $2x + 2y \frac{dy}{dx} = 0$ and $\frac{dy}{dx} = 2x$ , respectively. Solving this system of equations, we get the solutions $y = -\frac{1}{2}$ and $x = \sqrt{\frac{199}{2}}$ . Thus, $r^2 = x^2 + y^2 = \frac{399}{4}$ . The area of the circle is $r^2 \pi = \frac{399}{4} \times \pi$ , so $a + b = 399 + 4 = 403$ .

The radius of the circle we're after is the shortest distance from the origin $(0,0)$ to the parabola $(t,t^2-100)$ . By the distance formula this is the infimum of $\sqrt{(t^2 + (t^2-100)^2)}$ $= \sqrt{t^4 - 199t^2 + 100^2}$ and completing the square this is $= \sqrt{(t^2-199/2)^2 + 100^2-(199/2)^2} = \sqrt{(t^2 - 199/2)^2 + (200^2-199^2)/4}.$ $= \sqrt{(t^2-199/2)^2 + 399/4}.$ This takes minimal value of $\sqrt{399/4}$ for $t = \pm\sqrt{199/2}$ and so the radius of the circle is $r = \sqrt{399/4}$ . Thus the area of the circle is $\pi r^2 = \frac{399}{4} \pi$ . The fraction $399/4$ is reduced and so the answer is $399+4 = \boxed{403}$ .

The equation of the circle is $x^2+y^2=r^2$ . Now The circle and the parabola have 2 common tangent points, which are reflected by each other through the line $0,y$ hence $y_1=y_2$ and $x_1=-x_2$ . The equation of the parabola is $y=x^2-100$ . So we get $y^2+y+100-r^2=0$ This equation has one and only root since $y_1=y_2$ , then $1-4(100-r^2)=0$ So $r^2=\frac{a}{b}=\frac{399}{4}$ .The answer is $403$

General formula of a circle: $x^{2} + y^{2} = r^{2}$

Differentiation of the general formula gives: $\frac{dy}{dx} = \frac{-x}{y}$

At the point the circle and the parabola touches: $\frac{dy}{dx}(Parabola)$ = $\frac{dy}{dx}(Circle)$

$2x=\frac{-x}{y}$

$y = \frac{-1}{2}$

Putting the value of y in the equation for parabola:

$\frac{-1}{2}=x^{2}-100$

$x=\sqrt{\frac{199}{2}}$

Now putting the value of x and y in the general equation of circle to derive radius r,

$\sqrt{\frac{199}{2}}^{2} + \frac{-1}{2}^{2} = r^{2}$ $r = \sqrt{\frac{399}{4}}$

Area of the circle= $PI \times r^{2}$

$(\sqrt{\frac{499}{4}})^2 \times PI$ $\frac{399}{4} \times PI$

x^2 + y^2 = r^2.

y = x^2 - 100

y + 100 + y^2 = r^2

d(y + 100 + y^2)/dy = 0

2y + 1 = 0, y = -0.5

r^2 = -0.5 + 100 + 0.25 = 99.75 = 399/4

the area of circle = π.r^2 = π. 399/4. a+b = 399+4=403

Let, the equation of the circle is x^2+y^2=r^2.........(1), where r is the radius of the circle. We have the equation of the parabola, i.e. y=x^2-100 \Rightarrow x^2=y+100........(2) We know, that the circle is inscribed in the parabola. So, the two equations must have two common solutions. By putting the value of x^2 of (2) equation in the (1) equation, we get, y^+y+(100-r^2)=0 \Rightarrow y= \frac{-1 \pm \sqrt{(-1)^2-4(100-r^2)}{2}. In this case, the parabola and the circle both are symmetric with respect to the 'Y'-axis, I mean if we reflect the part of the parabola and the circle which lies on the 1st and 4th quadrant, with respect to the 'Y'-axis, we shall get the parts which lies on the 2nd and 3rd quadrant. So, the ordinates of the two touching points of the parabola and the circle, will be same. So there must be only one value of y, so [(-1)^2-4(100-r^2)] must be zero, i.e. (-1)^2-4(100-r^2)=0 \Rightarrow 4r^2-400+1=0 \Rightarrow 4r^2=399 \Rightarrow r^2= \frac {399}{4} \Rightarrow \pi \cdot r^2 = \pi \cdot \frac {399}{4} r is the radius of the circle, so the area of the circle is \pi \cdot r^2, which is equal to \pi \cdot \frac {399}{4} So, here, a=399 and b=4; so a+b=399+4=403.[answer]

I used a little calculus: Let the point of tangency be $a, a^2 - 100$ . We know the slope of the radius at that point is the negative of the reciprocal of the slope of the tangent line, which by calculus is $2a$ , so our slope of the radius is $-\dfrac{1}{2a} = \dfrac{a^2 - 100}{a}$ . Solving gives $a^2 - 100 = \dfrac{-1}{2}$ so $a^2 = \dfrac{199}{2}$ . The square of the radius is $a^2 + (a^2 - 100)^2 = \dfrac{199}{2} + \dfrac{1}{4} = \dfrac{399}{4}$ , so our answer is $\boxed{403}$ . $\blacksquare$

Let the radius of circle is R

Then, equation of circle: x^2 + y^2 = R^2 ....(1)

Given, equation of parabola: y = x^2 -100 ....(2)

We will find the intersection point of circle and parabola

*a). Put x^2 = y + 100, in ... (1)

y^2 + y + (100 - R^2) =0

Differentiate it w.r.t. y

2*y+1=0, y=-1/2, y^2=1/4

b). Put y=x^2-100 in ...(1)

* x^4 - 199 (x^2) + (10000- R^2) = 0**

Differentiate it w.r.t. x

4x^3 - 398x = 0

x(4x^2 - 398) = 0

x^2=0, 398/4

Now we have x^2 = 398/4, y^2 = 1/4 , x=0 can not be the intersection point

use the equation of circle, x^2 + y^2 = R^2 = (398/4) + (1/4) = 399/4

Area of circle = pi * (R^2) = (399/4) pi = (a/b) pi

a=399, b= 4, a+b=403

Eqn of parabola:

y = x^2 - b; b > 0.

The radius of the inscribed circle centered at the origin is the minimum distance of the parabola from the origin. This minimum distance happens when square of the distance of the parabola from origin is min. This min happens when

f(x) = D^2(x,y) = x^2 + y^2 = x^2 + (x^2-b)^2 = b^2 - (2b-1) x^2 + x^4

is min. This point is

x^2 = b - 1/2, y = -1/2 => r^2 = x^2+y^2 = b - 1/4 = 399/4.

What we want is a circle of radius r that intersects the parabola at most two points, so the above conditions will be met.

The equation $x^{2}$ + $y^{2}$ = $r^{2}$ can be rewritten as y = $\sqrt{r^{2} - x^{2}}$

Therefore $\sqrt{r^{2} - x^{2}}$ = $x^{2}$ - 100. Squaring both sides will give us:

$r^{2}$ - $x^{2}$ = $x^{4}$ - 200 $x^{2}$ + 10000. We can rewrite the equation into:

$x^{4}$ -199 $x^{2}$ + (10000- $r^{2}$ ) = 0. We should let the discriminant be zero for us to find the value of r which will only give two solutions.

So $b^{2}$ - 4ac = 0 ; $(-199)^{2}$ - $4 \times (10000 - r^{2})$

Then $r^{2}$ = 10000 - $\frac{(-199)^{2}}{4}$ ; $r^{2}$ = $\frac{40000 - 39601}{4}$ ; $r^{2}$ = $\frac{399}{4}$

Therefore the area of the circle is πr^{2} = π $\frac{399}{4}$ ; a + b = 403

For maximum the circle should exactly touch the parabola. So now substituting:- y=x^2 -100 in x^2 +y^2=a^2 ,where a is the radius of the supposed circle.

We get:- y+100+y^2=a^2

Solving the discriminant we get

[1-4(100-a^2)]^1/2>=0

Solving:- a^2=399/4

Area of circle=π a^2=π 399/4 Which gives us a=399,b=4 So a+b=399+4=403

Since the circle centered at the origin, then the equation of circle is x^2+y^2=r^2. and hence slope at any point will be = -x/y & slope of parabola will be equal to = 2x Since circle is inscribed in parabola slope at touching point will be same which gives y=-1/2 x^2=100.25 &r^2=100.5 area=π100.5=π399/4 so a+b=403

Let $P(a,b)$ be any point on the Cartesian plane. We define $D(P)$ to be the distance from origin to the point $P$ .

Claim:- If the circle touches the parabola at a point $T$ , then $T$ must the point on the parabola where $d(T)$ is minimized

*Proof:- * On the contrary assume not. Let $Q$ be the point on the parabola where $d(Q)$ is minimized. Since $T \neq Q$ , we know that there will be infinitely many points between $P$ and $Q$ ( $P$ is the point on the parabola where $d(P)$ is minimized). Let $M$ be any such point. Then $d(M) < d(Q)$ . However since $d(Q)$ is the radius of the circle, we find out that the circumference of the circle will lie outside point $M$ . This contradicts the fact that the circle is inscribed in the parabola. Hence if the circle touches the circle at a point $P$ , $d(P)$ must be minimized at that point.

Let $(a,b)$ be a point on the parabola where $D(a,b)$ , i.e $\sqrt{a^2 + b^2}$ is minimized. Since $a^2= b + 100$ , $D(a,b) = \sqrt{b^2 + b +100}$ . So we have to minimize $b^2 + b + 100$ . Note that $b^2 + b + 100= (b+\frac{1}{2})^2 + 100 - \frac{1}{4}$ . So the minimum value will be attained when $b= -\frac{1}{2}$ , and the minimum value is $\sqrt{100 - 1/4} = \sqrt{399/4}$ . So the radius of the circle is $\sqrt{399/4}$ . Therefore area= $\frac{399}{4} \pi$ . $a+b= 399+4= 403$

The equations for the parabola and the circle (respectively) are $y = x^2 - 100$ $x^2 + y^2 = r^2$ We can substitute $r^2 - y^2$ for $x^2$ to get the equation $y = r^2 - y^2 - 100$ which can be rearranged to $y^2 + y + 100 - r^2 = 0$ a simple quadratic equation. Since the circle is inscribed in the parabola, there can be only one value of y at which both curves intersect, therefore, the quadratic equation must be a square, meaning that $100 - r^2 = \dfrac{1}{4}$ We get $r^2 = \dfrac{399}{4}$ therefore $\pi r^2 = \dfrac{399 \pi}{4}$ and a + b = 403.

Because the circle is inscribed in the parabola, the tangential point must near the origin more than any point in the parabola.

This distance is also the radius r of the circle: $r^2=x^2+y^2$ and r min.

Because $y=x^2-100$ , so $r^2=y^2+y+100$ .

Clearly, r min when $y=\frac{-1}{2}$ . Therefore, $r^2=\frac{399}{4}$ and the answer is 399+4=403