JEE Main 2023 — Circle Question with Solution

Equation of AB :

$$y-(-5)=\frac{-5+11}{8-4}(x-8)$$

$$\Rightarrow y+5=\frac{3}{2}(x-8)$$

$$\Rightarrow 2y+10=3x-24$$

$$\Rightarrow 3x-2y-34=0$$ .... (1)

Also AB is cord of contact. And we know equation of cord of contact to a circle is $$T=0$$

$$\Rightarrow xh+yk-\frac{3}{2}(x+h)+5(y+k)-15=0$$

$$\Rightarrow x\left(h-\frac{3}{2}\right)+y(k+5)+\left(-\frac{3}{2}h+5k-15\right)=0$$ .... (2)

Comparing equation (1) and (2), we get

$$\frac{h-\frac{3}{2}}{3}=\frac{k+5}{-2}=\frac{-\frac{3}{2}h+5k-15}{-34}$$

$$\therefore$$ $$-2h+3=3k+15$$

$$\Rightarrow 3k+2h=-12$$ ..... (3)

and

$$17(k+5)=-\frac{3}{2}h+5k-15$$

$$\Rightarrow 12k+\frac{3}{2}h-100$$

$$\Rightarrow 3h+24k=-200$$ ..... (4)

Solving (3) and (4), we get

$$h=8$$ and $$k=-\frac{28}{3}$$

$$\therefore$$ Point C is $$\left(8,-\frac{28}{3}\right)$$

Now radius of the circle whose centre is at C and tangent is AB is

$$=\left|\frac{3(8)-2\left(-\frac{28}{3}\right)-34}{\sqrt{3^2+2^2}}\right|$$

$$=\left|\frac{26}{3\sqrt{13}}\right|$$

$$=\frac{2\sqrt{13}}{3}$$