Question

Prove that the angle between two tangents from the origin to the circle $(x-7{)}^2+(y+1{)}^2=25$ is $\pi /2$.

Answer 1

Any line through $(0,0)$ is $y-mx=0$
If it is a tangent to a given circle then applying the condition of tangency, i.e., $p=r$ where centre is $(7,-1)$ and $r=5$, we get
$\therefore \frac{-1-7m}{\sqrt{(m^2+1)}}=5$
or $(1+7m{)}^2=25(1+m^2)$
or $24m^2+14m-24=0$
Above gives us the slopes of the two tangents drawn from $(0,0)$.
Since $m_1{m}_2=-\frac{24}{24}=-1$.