JEE Main 2023 — Circle Question with Solution

Given,

The tangents at the points $A(4,-11)$ and $B(8,-5)$ on the circle $x^2+y^2-3x+10y-15=0$, intersect at the point $C$. 

So, equation of tangent at $A(4,-11)$ on circle will be

$\Rightarrow 4x-11y-3\left(\frac{x+4}{2}\right)+10\left(\frac{y-11}{2}\right)-15=0$

$\Rightarrow 5\mathrm{x}-12\mathrm{y}-152=0 .........\left(1\right)$

And equation of tangent at $\mathrm{B}(8,-5)$ on circle is

$\Rightarrow 8x-5y-3\left(\frac{x+8}{2}\right)+10\left(\frac{y-5}{2}\right)-15=0$

$\Rightarrow 13\mathrm{x}-104=0\Rightarrow \mathrm{x}=8 .........\left(2\right)$

Now putting the value of $x=8$ in equation $\left(1\right)$ we get, $\mathrm{y}=\frac{-28}{3}$

So, the point $C$ will be $C\left(8,\frac{-28}{3}\right)$

Now equation of tangent $AB$ will be,

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

$\Rightarrow 3x-2y-34=0$

$\mathrm{r}=\left|\frac{3\times 8+\frac{2\times 28}{3}-34}{\sqrt{13}}\right|=\left|\frac{-26}{3\sqrt{13}}\right|=\frac{2\sqrt{13}}{3}$