Coordinate Geometry - Circle

Find $\ell$ so that the point dividing the joint of $\left(4,7\right)$ and $\left(-5,6\right)$ in the ratio $\ell : 1$ should lie on the circle $x^2 + y^2 = 65$.

Let the given points be $A \left(4,7\right)$ and $B \left(-5, 6\right)$

Let $P \left(p, q\right)$ be the point dividing the join of points $A$ and $B$ in the ratio $\ell : 1$

Then, by section formula for internal division

$p = \dfrac{-5 \ell + 4}{\ell + 1}$ $\;$ and $\;$ $q = \dfrac{6 \ell + 7}{\ell + 1}$

Equation of the given circle is $\;$ $x^2 + y^2 = 65$

Since $P \left(p, q\right)$ lies on the given circle

$p^2 + q^2 = 65$

i.e. $\;$ $\left(\dfrac{-5 \ell + 4}{\ell + 1}\right)^2 + \left(\dfrac{6 \ell + 7}{\ell + 1}\right)^2 = 65$

i.e. $\;$ $25 \ell ^2 - 40 \ell + 16 + 36 \ell ^ 2 + 84 \ell + 49 = 65 \ell ^2 + 130 \ell + 65$

i.e. $\;$ $4 \ell ^2 + 86 \ell = 0$

i.e. $\;$ $2 \ell ^2 + 43 \ell = 0$

i.e. $\;$ $\ell \left(2 \ell + 43\right) = 0$

i.e. $\;$ $\ell = 0$ $\;$ or $2 \ell + 43 = 0$

$\because \;$ $\ell = 0$ is not possible for the given internal division

$\implies$ $2 \ell + 43 = 0$

$\implies$ $\ell = \dfrac{-43}{2}$