Question

Let tangents are drawn at two points of the circle $(x-7{)}^2+(y+1{)}^2=25$. If the point of intersection of both the tangents is origin, then the angle between them (in degrees) is

Answer 1

Any line passing through origin is written as
$y-mx=0$
This line is tangent to the circle $(x-7{)}^2+(y+1{)}^2=25$, so the distance from centre to the line is equal to radius,
$\frac{|-1-7m|}{\sqrt{1+m^2}}=5$
Squaring on both the sides, we get
$(1+7m{)}^2=25(1+m^2)$
$\Rightarrow 24m^2+14m-24=0D={14}^2+4\times {24}^2>0$
Let the roots of the equation be $m_1,m_2$, so
$m_1{m}_2=-\frac{24}{24}=-1$

Hence, the angle between the tangents is ${90}^{\circ }$


Alternate solution :
Given circle is $(x-7{)}^2+(y+1{)}^2=25$


Distance between $OC$
$=\sqrt{7^2+1^2}=\sqrt{50}=5\sqrt{2}$
From the figure, we get
$\sin \frac{\theta }{2}=\frac{CP}{OC}\Rightarrow \sin \frac{\theta }{2}=\frac{5}{5\sqrt{2}}=\frac{1}{\sqrt{2}}\Rightarrow \frac{\theta }{2}={45}^{\circ }\therefore \theta ={90}^{\circ }$