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Lines 5x + 12y - 10 = 0 and 5x - 12y - 4...

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

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According to the question, circles are drawn as shown in the figure.

Centre of the circles lies on one of the angle bisectors of the given lines.
`(5x+12y-10)/(13)= +- (5x-12y-40)/(13)`
or `24y=-30` and `10x=50`
But centre lies in first quadrant.
So, for centre , `x=5`.
Distance of centre (C(5,y) from the line `5x+12y-10=0` is 3.
`:. (5(5)+12y-10)/(13)= +-13`
`:. y=2` (As centre lies in first quadrant)
Sol, centre is (5,2).
Alsok, in right angles triangle CMR, `CR=5(` as `RM=4)`.
Hence, equation of circle is `(x-5)^(2)-(y-2)^(2)=5^(2)`.
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