Circle Ques 108
- Lines $5 x+12 y-10=0$ and $5 x-12 y-40=0$ touch a circle $C _1$ of diameter 6 . If the centre of $C _1$ lies in the first quadrant, find the equation of the circle $C _2$ which is concentric with $C _1$ and cuts intercepts of length 8 on these lines.
(1986, 5M)
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Answer:
Correct Answer: 108.$(x-5)^{2}+(y-2)^{2}=5^{2}$
Solution:
Formula:
- Since, $5 x+12 y-10=0$ and $5 x-12 y-40=0$ are both perpendicular tangents to the circle $C _1$.
$\therefore O A B C$ forms a square.
Let the centre coordinates be $(h, k)$, where,
$$ \begin{gathered} O C=O A=3 \text { and } O B=6 \sqrt{2} \\ \Rightarrow \quad \frac{|5 h+12 k-10|}{13}=3 \text { and } \frac{|5 h+12 k-40|}{13}=3 \end{gathered} $$
$\Rightarrow \quad 5 h+12 k-10= \pm 39$ and $5 h-12 k-40= \pm 39$ on solving above equations. The coordinates which lie in I quadrant are $(5,2)$.
$\therefore$ Centre for $C _1(5,2)$
To obtain equation of circle concentric with $C _1$ and making an intercept of length 8 on $5 x+12 y=10$ and $5 x-12 y=40$ $\therefore$ Required equation of circle $C _2$ has centre $(5,2)$ and radius 5 is $(x-5)^{2}+(y-2)^{2}=5^{2}$
BETA
