The absolute difference between the squares of the radii of the two circles passing through the point $(-9,4)$ and touching the lines $x+y=3$ and $x-y=3$, is equal to ______.

Question

Aicharya · Accepted Answer

Centre $(\mathrm{a}, 0)$
$\mathrm{r}=\left|\frac{\mathrm{a}-0-3}{\sqrt{2}}\right|$
$\operatorname{circle}(x-a)^2+y^2=\left(\frac{a-3}{\sqrt{2}}\right)^2$
passes through $(-9,4)$
$2\left(a^2+18 a+81+16\right)=\left(a^2-6 a+9\right)$
$\mathrm{a}^2+42 \mathrm{a}+185=0$
$(a+37)(a+5)=0$
$\Rightarrow a=-37,-5$
$r_1=\left|\frac{-37-3}{\sqrt{2}}\right|=20 \sqrt{2}$
$\mathrm{r}_2=\left|\frac{-5-3}{\sqrt{2}}\right|=4 \sqrt{2}$
$\left|r_1^2-r_2^2\right|=|800-32|=768$

The absolute difference between the squares of the radii of the two circles passing through the point $(-9…

Solution