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Explanation

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)`