IIT-JEE 1981Question 1
Find the equations of the circles passing through $(-4,3)$ and touching the lines $x+y=2$ and $x-y=2$.
Circle
Answer$x^{2}+y^{2}+2\left(10\mp\sqrt{54}\right)x+55\mp8\sqrt{54}=0$
Solution
Concept. The centre lies on the bisector of the angle that contains $(-4,3)$. The radius is the distance from the centre to either line. Formulas used. (1) Bisectors of $x+y-2=0$, $x-y-2=0$: $y=0$ and $x=2$.(2) Distance of $(h,0)$ from $x+y-2=0$: $\frac{|h-2|}{\sqrt2}$.Step 1 - Which angle. At $(-4,3)$: $x+y-2=-3$ and $x-y-2=-9$, same sign. Points $(h,0)$ with $h<2$ also give the same sign. So the centre is $(h,0)$, $h<2$. Step 2 - Use (2). $r=\frac{2-h}{\sqrt2}$. Step 3 - Through $(-4,3)$. $$(h+4)^2+9=\frac{(2-h)^2}{2}\Rightarrow2h^2+16h+50=h^2-4h+4\Rightarrow h^2+20h+46=0.$$ $h=-10\pm\sqrt{54}$ (both $<2$).Step 4 - Equation. $x^2+y^2-2hx+(h^2-r^2)=0$, where $$h^2-r^2=\frac{h^2+4h-4}{2}=\frac{(-20h-46)+4h-4}{2}=-8h-25.$$ For $h=-10\pm\sqrt{54}$: $-2h=2(10\mp\sqrt{54})$ and $-8h-25=55\mp8\sqrt{54}$. Answer: $x^{2}+y^{2}+2\left(10\mp\sqrt{54}\right)x+55\mp8\sqrt{54}=0$