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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, find the equation of the circle C2 which is concentric with C1 and cuts intercepts of length 8 on these lines.

A
x2+y210x4y+4=0
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B
x2+y2+10x4y+4=0
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C
x2+y210x+4y6=0
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D
x2+y210x+4y+6=0
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Solution

The correct option is C x2+y210x4y+4=0
Since the lines 5x+12y10=0 and 5x12y40=0 touches the circle C1,
its center lies on one of the angle bisector of the given lines, i.e. on the lines
5x+12y1025+144=±5x12y4025+1445x+12y10=±5x12y40
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+12y10=0 touches C1 we have
5×5+12t1025+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+42r=5
Hence the equation of C2 is
(x5)2+(y2)2=52x2+y210x4y+4=0

389334_257525_ans_dab2fb9c9e0044a68e24772b8abbc7b2.png

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