Radius is rational

Add a point $E$ inside the circle so that $ABCE$ is a rectangle. Adding this point gets us: $AE = BC = 15$ , $CE = AB = 12$ and since $\angle DCB = \angle ECB = 90°$ , $D$ , $C$ and $E$ are collinear which gets us: $DE = CD + EC = 8 + 12 = 20$ and $AD = \sqrt{AE^2 + DE^2} = 25$ . Also, $BD = \sqrt{BC^2 + CD^2} = 17$ .

Now, extend the line $BC$ until it cuts the circle at point $F$ . Because $\angle ABF = \angle ABC = 90°$ , $AF$ is the diameter of the circle ( $AF = 2R$ ). $ABDF$ is a cyclic quadrilateral, so: $\angle ADF = \angle ABF = 90° = \angle BCD$ and $\angle DBF = \angle DAF$ . This yields that $\triangle BCD$ and $\triangle ADF$ are similar, so: $\dfrac{AF}{AD} = \dfrac{2R}{AD} = \dfrac{BD}{BC}$ .

Finally, the radius of the circle is: $R = \dfrac{AD*BD}{2*BC} = \dfrac{25*17}{2*15} = \dfrac{85}{6} \boxed{\approx 14.167}$

BC = 17, by Pythagorean theorem

AC = 25, by Pythagorean theorem ((AB+CD)^2 + BD^2 = AC^2)

The radius R of the circumcircle of triangle ABC with a given sides a,b,c is:

R = $\frac{abc}{√(a+b+c)(b+c-a)(c+a-b)(a+b-c)}$ = $\frac{12*17*25}{√54*30*20*4}$ = $\frac{5100}{360}$ =14 $\frac{1}{6}$

We have to set $BC=12, CD=15, DE=8, DF=x$ , for similar triangle $\Delta DEG\sim \Delta CBG$ , we have that $\dfrac{GD}{CG}=\dfrac8{12}=\dfrac23$

Hence $GD=6, CG=9$ .

Also, there is a theorem stats that $FG\cdot CG=EG\cdot GB$ but I not sure about the name, hence $9(x+6)=150\rightarrow x=\dfrac{32}3$

Then the diameter is $2R$ where $R$ stated as radius, so $4R^2=CF^2+BC^2$

which give us $R=\boxed{\dfrac{85}{6}}$ .

$r^2=(15-\sqrt{r^2-6^2})^2+14^2$ $r^2=(225-30\sqrt{r^2-36}+(r^2-36))+196$ $r^2=385-30\sqrt{r^2-36}+r^2$ $30\sqrt{r^2-36}=385$ $r^2=\frac{385}{30}^2+36=\frac{7225}{36}$ $r=\sqrt{\frac{7225}{36}}=\frac{85}{6}\approx \boxed{14.167}$

In circle , $x * 15 = 8 * 20$ , so $x = \dfrac{32}{ 3 }$

and $( 2R)^2 = 12^2 + (15+x) ^2$

So we can find $R = 14.166666666666666666666666666667$

Extend $\overline{BC}$ to $\overline{BCE}$ where $E$ lies on the circle, and let $P$ denote the intersection of $\overline{BC}$ and $\overline{AD}.$

Then $\triangle{ABP}, \triangle{DCP}$ are similar right triangles, from which we can conclude $BP = 9, PC = 6, AP = 15, PD = 10$

Now, using the power of the point $P$ , we see $150 = AP \cdot PD = BP \cdot PE = 9 \cdot (6+CE) \implies CE = \frac{32}{3}$

Lastly, since $\triangle ABE$ is a right triangle inscribed in the given circle, its hypotenuse is a diameter of the circle, so the radius is $\frac{1}{2}AE = \frac{1}{2}\sqrt{12^2 + \left(15 + \frac{32}{3}\right)^2} = \frac{1}{2}\left(\frac{85}{3}\right) = \frac{85}{6} \approx \boxed{14.16667}$

Please see as follow chart, we could regard point A,B,C on cirle is to be define as X,Y coordinate ,for example we define O(x.y) is coordinate of the center of cirle Meanwhile the point A is (0,0) as Origin of coordinate system

Step 1: since OA = OB = OC = R then:

         x² + y² = ײ+ (y-12)²
         x² + y²  = (x-15)² +(y-29)²

Step 2: it is easy to Solve equations x = 77/6 y=6

Step 3: R**2 = x² + y²

R = 14.16666... - \

Let E be the mid-point of AB have coordinates (-x,0),the circle center O be (0,0), Then A is (-x,6) and D = (15 -x,14). Then r^2 = (OA)^2 = (OD)^2. So X^2 + 36 = (x - 15)^2 + 196, or 30x = 385, and x = 77/6. Then r^2 = (77/6)^2 + 36 = 7225/36, and r = 85/6 = 14.167. Ed Gray

The centre of the circle lies at the intersection of axes $AB$ and $BC$ .

Let's start coordinate system in the middle of $BC$ to simplify equation for $q$ , then:

$p: y=-10$

$q: y=-\frac{15}{8}x$

$\rightarrow$ $S [x=\frac{16}{3}, y=-10]$

$r = |\vec{BS}| = |\vec{BE}+\vec{ES}| = |(7.5+\frac{16}{3}, 4-10)| = \sqrt{(7.5 + \frac{16}{3})^2 + (-6)^2}$ $\boxed{ \approx 14.167}$

Hope it's clear enough..

I just did this:

1. The 12 cm line is centered, so I can find the center of the circle by dividing it in half.

   $\frac{12}{2}$ = 6

2. The 8 cm line added to the previous result create something similar to the radius of the circle.

   6+8 = 14

This isn't a exact method, but it works if you want to solve it quickly.

I USED ACAD

$Let\quad a=BC,\quad b=AC\quad and\quad c=AB\quad ,R\quad the\quad radius\quad of\quad the\quad circle\quad and\quad A\quad its\quad area.\quad \\ We\quad have\quad R=\frac { abc }{ 4A } \quad \\ a=\sqrt { 8²+15² } =17\quad b=\sqrt { 15²+20² } =25\quad c=12\\ Let\quad h=\frac { a+b+c }{ 2 } \\ Using\quad the\quad Heron\quad of\quad alexandri\quad \\ A=\sqrt { h(h-a)(h-b)(h-c) } =90\\ Then\quad R=\frac { 85 }{ 6 } =14.167$

The three points A, B and C fulfill the circle equation

$(x-x_0)^2+(y-y_0)^2=r^2$

and A is the origin of the coordinate system. This leads to three equations:

(1) $x_0^2+y_0^2=r^2$

(2) $x_0^2+(12-y_0)^2=r^2$

(3) $(15-x_0)^2+(20-y_0)^2=r^2$ .

Solving this system gives the result: $y_0=6$ , $x_0=\frac{77}{6}$ and $\boxed{r=\frac{85}{6} \approx 14.167}$ .

We have 3 points on the circle as (0,0) , (0,12) and (15,20). The circle's equation is (x-a)^2 + (y-b)^2 = r^2. Substituting for the 3 points, we get a^2 + b^2 = r^2 a^2 + (12-b)^2 = r^2 (15-a)^2 + (20-b)^2 = r^2 We can solve these for a = 77/6, b = 6. This yields r = sqrt(a^2+b^2) = 14.167

I made a solution using functions. The equation for a circle with radius r in cartesian plain is given by $r^2=(x-a)^2+(y-b)^2$ , where the values of $a$ and $b$ depend on the position of the circle retative to the plain.

We can choose any place to be the origin, so I'll put it on the spot marked on the image. The point (15,20) has its y value because the red and green lines are parallel (since the blue one is perpendicular to both) and add up to 20.

We can plug in the x and y values that appear on the circle in order to isolate $a$ $x=0$ and $y=0$

$r^2=(0-a)^2+(0-b)^2$

$r^2=a^2+b^2$

$r^2-b^2=a^2$

Now, we can use $x=0$ and $y=12$ $r^2=a^2+(12-b)^2$

We can substitute $a^2$ by the expression we found earlier: $r^2=r^2-b^2+144-24b+b^2$

$0=144-24b$

$b=\frac{144}{12}$

$b=6$

Finally, we can use $x=15$ and $y=20$

$r^2=(15-a)^2+(20-6)^2$

Using the first equation, we can substitute $r$ .

$a^2+6^2=225-30a+a^2+14^2$

$30a=385$

$a=\frac {77}{6}$

Finally, we use the first equation again to get: $r^2= ( \frac{77}{6})^2+6^2$

$r=sqrt( \frac{7225}{36})$

$r=\frac {85}{2}$

First we construct the lines i and j.

Then, we find i using the Pythagorean Theorem. i.e. $i=sqrt(15^2+20^2)=25$

Angle $α$ can be calculated using the fact that $tan(α)=8/15$

We can see that $90+α=δ/2=(360-β)/2$ , so $β=360-2(α+90)$

Knowing $i$ and angle $β$ we can determine the Radius.

Using the figure, set up two equations for X and R

X^2+ (6)^2= R^2

(15-X)^2 + (14)^2 = R^2

Whose, solution gives value of R as Answer = (85/6)=14.167

Let the red edge have vertices $AB$ , the blue edge have vertices be $BC$ , and the green vertices be $CD$ . Suppose $A=(0,-12)$ , $B=(0,0)$ , $C=(15,0)$ , $D=(15,8)$ .

The centre of the circle through $ABD$ is the intersection of the perpendicular bisectors of $AB$ and $BD$ . The first bisector is simply $y=-6$ . The second bisector is $y = - \frac{15}{8} x + \frac{289}{16}$ by a quick calculation.

Thus, the centre of the circle is $(\frac{77}{6}, -6)$ , which is a distance of $\frac{85}{6}$ from $B$ . Thus, the radius is $\frac{85}{6}$ .

$a=12 , b=15 , c= 8$
$r =\frac{\sqrt{(b^{2}+c^{2})((a^{2}+c^{2})+b^{2})}}{2b}= \frac{85}{6} =14.1666$

Half of 12 is 6, +8 is 14.

Its much easier: Just take the half of the line with 12 and add the parallel line of 8, so you have 6 + 8 = 14

AB is bisected by the equator of the circle. So, the bottom of the circle is a reflection of the top. 8+12+8=28=diameter Radius =0.5D R=14.0 to 3 sig figs which means the last digit is uncertain but is within a magnitude of the true number.

I literally just guessed. I'm a flippin genius 😊

Apologies for lack of LateX but ...

Take the point where the 12 and 15 length points to be the origin. We have the 12 length line is a chord of the circle. Join the origin to the end of the length 8 line to form a second chord of the circle. The centre of the circle lies on a perpendicular bisector of the chord. Thus the centre of the circle lies at the intersection of the perpendicular bisectors of the chords we have defined.

The first chord is just defined as y = -6. For the other one, the chord has gradient 8/15 so the perpendicular bisector has gradient -15/8 and passes through (15/2, 4). Thus this line is defined by y = (-15/8)x + c and we solve for c by putting in the intersection point to get 4 = (-15/8)15/2 + c. This gives c = 289/16

Solving for the intersection of perpendicular bisectors gives y = -6 and -6 = (-15/8)x + 289/16 which gives x = 77/6.

The radius of the circle is then just the distance from this centre to the origin = sqrt(6^2 + (77/6)^2) = 85/6.

(12+8+8)/2=14

Lengths in the diagram were found using Pythagoras' Theorem. $\angle O$ was found using the inscribed angle theorem.

Use the law of cosines on $\triangle ABC$ :
$\begin{aligned} 17^2 &= 12^2 + 25^2 - 2 \cdot12\cdot25 \cos \alpha \\ \cos\alpha &= \frac 45 \end{aligned}$

Use the law of cosines on $\triangle OAB$ :
$\begin{aligned} 17^2 &= 2r^2 - 2r^2 \cos 2\alpha \\ &= 4r^2 ( 1 - \cos ^2 \alpha ) \\ r &= \frac{85}6 \approx 14.167 \end{aligned}$

I just did some algebra. The center is at intersection between perpendiculars to middle points of segments AB and BD (from Novak's drawing). Let (0,0) be the middle point of segment AB. The perpendicular to middle point of segment AB is just the "y=0" line. The middle point of segment BD is (7.5,10), i.e. x=BC/2 and y=6+CD/2 where 6 is y-coordinate of B or C. The perpendicular to middle point of segment BD is y = -15/8*(x-15/2)+10 where the line slope is the opposite inverse of that of the segment BD. Then, the center is located at x=16/3+15/2 (and y=0). The radius is just the distance between the center and point B, i.e. sqrt(6^2 + (16/3+15/2)^2)=14.1667.

Let's call the points given on the circle $A, B, C$ going clockwise from the bottom. We want to find the radius of the circle, which is the circumscribed circle of $\triangle ABC$ . This means we need to find the center, which we can find at the intersection of the perpendicular bisectors of a triangle. Let's add a coordinate system. Since we're not restricted in our choice of coordinates, we just pick a convenient one: let our X-axis be the perpendicular bisector of $AB$ . Then $A=(0,-6)$ , $B=(0,6)$ , $C=(15,14)$ .

The midpoint $D$ of $BC$ is $(7.5,10)$ , and the vector $B-D$ is $(7.5,4)$ . Rotation by 90 degrees clockwise means swapping the coordinates and then negating the second, giving us a vector $V=(4,-7.5)$ . The points which can be expressed as $D+t\cdot V$ for real numbers $t$ are precisely the points on the perpendicular bisector on $BC$ (by anchor point $D$ being the midpoint and direction vector $V$ being perpendicular to $BC$ ).

Since we want an intersection of two perpendicular bisectors, with our X-axis being one and the set of points given by $D+t\cdot V$ (that is, $(7.5,10) + t \cdot (4, -7.5)$ ) for real $t$ being another, let's find a $t$ which puts such a point on the X-axis. In other words, the second coordinate needs to be $0$ .

$10 + t \cdot (-7.5) = 0$ gives us $t =\frac{4}{3}$ . Calculating the first coordinate from that $t$ yields $7.5 + \frac{4}{3} \cdot 4 = 12 \frac{5}{6}$ . So our midpoint $M$ should be at $(12\frac{5}{6},0)$ . Calculate distances to $B$ and $C$ by Pythagoras (the distance to $A$ is equal to the distance to $B$ by symmetry across the X-axis), and check if they're the same: both work out to $14.16\overline{6}$ , so $M$ is indeed the midpoint of the circle and the distance found is the distance sought, which after rounding works out to $14.167$ .

8 + 12 + 8 = 28 ------> 28/2 = 14. The answer simplified.

The perpendicular bisector of a chord (the red line) passes through the center of the circle, a diameter. Half the red chord (6) plus the green line (8) is a radius = 14.

This can be solved just by introducing a coordinate frame with A(0, 0), than B(0, 12) and C(15, (12+8) ). Than the center of the circle has coordinate $O(x_{R}, y_{R})$ . We know that |OA| = |OB| = |OC|, i.e with the chosen coordinate frame: $x_{R}^2+y_{R}^2 = x_{R}^2+(y_R-12)^2 = (x_R-15)^2+(y_R-20)^2$ . From the first two equations we can see that $y_R = 6$ , and from the last two $x_R = 77/6$ .

Using the points $P$ , $Q$ , and $R$ on the circle above $\implies$

$a^2 + b^2 = r^2$

$(15 - a)^2 + (8 - b)^2 = r^2$

$a^2 + (12 + b)^2 = r^2$

Solving the nonlinear system above we obtain:

$b = -\dfrac{144}{24} = -6$ , $a = \dfrac{385}{30} = \dfrac{77}{6} \implies r^2 = \dfrac{7225}{36} = \dfrac{85^2}{6^2} \implies r = \dfrac{85}{6} \approx \boxed{14.167}$ .

For me using coordinate geometry is the simplest approach.

Fermi estimation: 8+(12/2)=14

This is not an exact solution but gives a close approximation relying on the accuracy of the pictorial representation. Because the 12 length is a vertical and tangential line, half it's length partially makes up the total length of the radius. It is assumed that the 15 length line is horizontal because of the right angle with the 12 length, and the same is said for the 8 length being vertical.

It is discernable that the 8 length is not coincident with the center, but by adding it's length to half 12, we have a simple Fermi estimation that can be used to check against the results of trigonometric problem solving.

Again, 14 is not an exact answer for the radius, but it satisfies the margin of error.