Suppose the tangents drawn to the circle $x^2+y^2-6 x-4 y-11=0$ from $P(1,8)$ touch the circle at $A$ and…
Suppose the tangents drawn to the circle $x^2+y^2-6 x-4 y-11=0$ from $P(1,8)$ touch the circle at $A$ and $B$. Then the centre of the circle passing through $P$, $\mathrm{A}$ and $\mathrm{B}$ is
$(2,5)$
$(-2,-5)$
$(-2,5)$
$(2,-5)$
Solution
Given equation of circle is $x^2+y^2-6 x-4 y-11=0$
$
\therefore \text { centre }=(3,2)
$
A circle passing through $\mathrm{P}, \mathrm{A}$ and $\mathrm{B}$ then $\mathrm{PC}$ is a diameter
$\therefore$ centre of required circle is $\left(\frac{1+3}{2}, \frac{8+2}{2}\right)$ i.e. $(2,5)$