Question

A variable circle always touches the line $y-x$ and passes through the point $(0,0)$. The common chords of above circle and $x^2+y^2+6x+8y-7=0$ will pass through a fixed point whose coordinates are

• A. $(-\frac{1}{2},\frac{3}{2})$
• B. $(-\frac{1}{2},-\frac{1}{2})$
• C. $(\frac{1}{2},\frac{1}{2})$
• D. None of these

Answer 1

The correct option is A $(-\frac{1}{2},\frac{3}{2})$

Equation of the line perpendicular to $x=y$ and through the point $(0,0)$ is $y+x=0$

if $x-$coordinate of centre$=h$

then $y-$coordinate of ventre$=-h$

then radius of the circle$=$Perpendicular distance from $(h,-h)$ on $x-y=0$

$=\frac{2h}{\sqrt{2}}=h\sqrt{2}$

$\therefore$Equation of the circle is

${(x-h)}^2+{(y+h)}^2=2h^2$

$\Rightarrow x^2-2xh+h^2+y^2+2yh+h^2=2h^2$

$\Rightarrow x^2+y^2-2xh+2yh=0$ ...........$\left(1\right)$

and other circle is $x^2+y^2+6x+8y-7=0$ ......$\left(2\right)$

$\therefore$ Equation of common chord of equations $\left(1\right)$ and $\left(2\right)$ is

$x^2+y^2-2xh+2yh-x^2-y^2-6x-8y+7=0$

$\Rightarrow 2x(h+3)+2y(4-h)-7=0$

$\Rightarrow (6x+8y-7)+2h(x-y)=0$

$\therefore 6x+8y-7=0$ and $x-y=0$

$\Rightarrow 6x+8y=7$ and $x=y$

$\Rightarrow 6x+8x=7$ since $x=y$

$\Rightarrow 14x=7$

$\Rightarrow x=\frac{1}{2}$

$\therefore x=y=\frac{1}{2}$ since $x=y$

Then, coordinates of the fixed point is $(\frac{1}{2},\frac{1}{2})$