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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.

Daniel
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1 Answers1

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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, $tan({\theta\over 2})$ = $r\over {PQ}$

$\Rightarrow tan({53^{\circ}\over 2})$ = $1\over 2$ = $r\over {\sqrt {200}}$

$\Rightarrow r = \sqrt {50}$

Dhanush Krishna
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  • It's better not to say $\theta=53$, since that is a approximation. Just skip that part(since that is unecessary). – S.C.B. Mar 20 '16 at 14:28