Question

Find the equation of tangents at point $(5,12)$ and $(12,-5)$ at circle $x^2+y^2=169.$ Provethat they will be perpendicular to each other. Also find the co-ordinates of intersection point.

Answer 1

Equation of circle
$x^2+y^2=\mathrm{169...}\left(i\right)$
At point $(x_1,y_1)$ equations of tangent line passing through the circle $x^2+y^2=a^2$
$x{x}_1+y{y}_1=a^2$
At point $(5,12)$
$x\times 5+y\times 12=169$
$\Rightarrow 5x+12y-169=0$
$\Rightarrow$ Gradient of this line,
$m_2=\frac{12}{5}$
Line (ii) and (iii) will be ${\perp }^r$ to each other
If $m_{1}m-2=-1$
$\therefore m_1{m}_2=\frac{-5}{2}\times \frac{12}{5}$
$=-1$
Thus, line (ii) and (iii) are ${\perp }^r$ to each other
For intersection point, solving equation (ii) and (iii)
Eqn, (ii) $\times 5$ and eqn (iii) $\times 12$
$(5x+12y=169)\times 5$
$(12x-Sy=169)\times 12$
$\Rightarrow 25x+60y=845$
$144x-60y=2028$
$169x=2873$
$x=\frac{2873}{169}=17$
Put value of $x$ in eqn (ii)
$5\times 17+12y-169=0$
$85-169+12y=0$
$-84+12y=0$
$\Rightarrow y=\frac{84}{12}=7$
Thus, intersection point is $(17,7)$.