A circle having centre at C is made to pass through the point $P(1,2)$ , touching the straight lines $7x - y = 5$ and $x + y +13 = 0$ at A and B respectively. Then find the radius of the circle.
I have no clue how to solve this problem. Please help me.
First of all, the important thing is to note that point $P(1,2)$ lies on the line $L_1=7x-y=5$. So, point P is point A itself.
The point of intersection of the lines $L_1 and L_2$ is $Q(-1,-12)$. Distance between P and Q is $\sqrt {200}$ which can be seen from the distance formula.
Next we find the angle between the lines is $tan(\theta)$=$|{{m_1 - m_2} \over {1-m_1m_2}}|$ where $m_1$ = -1 and $m_2$ = 7. We get $tan(\theta)$=$4\over3$ $\Rightarrow \theta=53^{\circ}$.
Now, $tan({\theta\over 2})$ = $r\over {PQ}$
$\Rightarrow tan({53^{\circ}\over 2})$ = $1\over 2$ = $r\over {\sqrt {200}}$
$\Rightarrow r = \sqrt {50}$