A geometry problem by Christopher Boo

Circle $\large { x }^{ 2 }+{ y }^{ 2 }-2x+4y-4=0$ is intercepted by $y=x+b$ or $\large \frac { y-x }{ b } =1$

We manipulate the equation and write it as $\large { x }^{ 2 }+{ y }^{ 2 }-2x(1)+4y(1)-4{ (1) }^{ 2 }=0$ .

Now we homogenise equation of circle with the line $\large \frac { y-x }{ b } =1$ $\large { x }^{ 2 }+{ y }^{ 2 }-2x(\frac { y-x }{ b } )+4y(\frac { y-x }{ b } )-4{ (\frac { y-x }{ b } ) }^{ 2 }=0$

The equation that we have got is equation of pair of straight lines passing through the points of intersection of the circle and the line and the origin.

It is given that circle with $AB$ as diameter passes through origin, this means $OA$ and $OB$ are perpendicular to each other ( $O$ is the origin). Proof of this statement is trivial and is left as a proof to the reader.

As they are perpendicular, coefficient of ${x}^{2}$ +coefficient of ${y}^{2}$ =0.

$\large 1+1+\frac { 2 }{ b } +\frac { 4 }{ b } -\frac { 4 }{ { b }^{ 2 } } -\frac { 4 }{ { b }^{ 2 } } =0$

$b=1$ or $b=-4$ .

$\large \boxed {-3}$

Click here for more details on homogenisation.

The family of circles passing through the intersection of given circle and line can be represented by

$x^2+y^2-2x+4y-4+\lambda(x-y+b)=0$

where $\lambda$ is a parameter.

As the new circle has $AB$ as its diameter, the centre of the circle $\left(1-\dfrac{\lambda}{2}, \dfrac{\lambda}{2} -2\right)$ must lie on $AB$ . So,

$\dfrac{\lambda}{2}-2=1-\dfrac{\lambda}{2}+b$

Or $\lambda=3+b$

Also since the circle passes through the origin, the constant term must be zero.

So, $\lambda\cdot b=4$

Solving the two equations we get $b^2+3b-4=0$

Therefore the sum of possible values of $b$ is $-3$

Note that the equation of the circle can be rewritten as: $(x-1)^2+(y+2)^2=3^2$ So, the circle is centred at $(1,-2)$ .

Saying that AB is the diameter of the circle is equivalent to saying that AB passes through the centre of the circle. i.e. The line $y=x+b$ passes through $(1,-2)$ .

This happens only when $b=-3$ .

Let the three points be $O(0,0)$ , $A(x_1,y_1)$ , $B(x_2,y_2)$ .

Since $AB$ is the diameter of the new circle, then $AOB$ is a right triangle.

$\begin{aligned} AB^2 &= AO^2+OB^2 \\\\ (x_1-x_2)^2+(y_1-y_2)^2 &= x_1^2+y_1^2+x_2^2+y_2^2 \\\\ x_1x_2+y_1y_2 &= 0 \end{aligned}$

This gave us an idea to try to solve for $POR$ .

$f(n) = \left\{ \begin{array}{l l} x^2+y^2-2x+4y-4 = 0 \\ x-y+b=0 \end{array} \right.$

Substitute $x=y-b$ into the first equation,

$\begin{aligned} (y-b)^2+y^2-2(y-b)+4y-4 &= 0 \\\\ 2y^2+(2-2b)y+(b^2+2b-4) &= 0 \end{aligned}$

Hence,

$\begin{aligned} y_1+y_2 &= b-1 \\\\ y_1y_2 &= \frac{1}{2}(b^2+2b-4) \end{aligned}$

Substitute $y=x+b$ into the two equation,

$\begin{aligned} x_1+x_2&=-(b+1) \\\\ x_1x_2 &= \frac{1}{2}b^2+2b-2 \end{aligned}$

Substitute $x_1x_2$ and $y_1y_2$ into the very first equation,

$\begin{aligned} x_1 x_2+y_1 y_2 &= 0 \\\\ \frac{1}{2} b^2+2b-2 +\frac{1}{2}(b^2+2b-4) &= 0 \\\\ b^2+3b-4 &=0 \end{aligned}$

Sum of roots $b$ is $-4+1=-3$ .