its center lies on one of the angle bisector of the given lines, i.e. on the lines
5x+12y−10√25+144=±5x−12y−40√25+144⇒5x+12y−10=±5x−12y−40
That is x=5 or y=−54
Since the center of C1 lies in the first quadrant.
Let its coordinates be (5,t) where t>0
As 5x+12y−10=0 touches C1 we have
∣∣∣5×5+12t−10√25+144∣∣∣=3 (radius of C1)
⇒15+12t=±3×13=±39
That is t=2 or t=−92
Now, we must take t=2, so that the center of C1 is (5,2) in the first quadrant
If r is the radius of the circle C2, we have
r2=32+42⇒r=5
Hence the equation of C2 is
(x−5)2+(y−2)2=52⇒x2+y2−10x−4y+4=0