Question

Lines `5x + 12y - 10 = 0` and `5x - 12y - 40 = 0` touch a circle C1 of diameter 6. If the centre of C1, lies in the first quadrant then the equation of the circle C2, which is concentric with C1, and cuts intercepts of length 8 on these lines is

Answer 1

Since, ` 5x + 12y - 10 = 0 and 5x - 12y - 40 = 0` are both perpendicular tangents to the circle `C_(1)`.
`therefore` OABC forms a square.
Let the centre coordinates be ((h, k), where,
`OC=OA=3 and OB = 6sqrt2`
`rArr(|5h+12k-10|)/(13)=3and (|5h+12k-40|)/(13)=3`

`rArr 5h + 12k - 10 = pm 39 and 5h - 12k - 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 5x + 12y = 10 and 5x - 12y = 40 `therefore` Required equation of circle `C_(2)` has centre (5, 2) and radius 5 is `(x-5)^(2) + (y-2)^(2)=5^(2)`