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Question

The center of a circle which passes through the points (0,0),(1,0) and touches the circles x2+y2=9 is:

A
(32,12)
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B
(12,32)
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C
(12,12)
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D
(12,2)
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Solution

The correct option is A (32,12)
Let the points P(0,0) and Q(1,0) touches the circle x2+y2=r2

Equation of circle is given by (xh)2+(yk)2=r2(1)

consider point P(0,0), we get
(0h)2+(0k)2=r2
h2+k2=r2(2)

consider point Q(1,0), we get
(1h)2+(0k)2=r2
1+h22h+k2=r2(3)

from equations (2) and (3), we get

h2+k2=1+h22h+k2

12h=0

h=12

Since, circle (1) touches circle (2), then the center is given by center of

circle (1)=(h,k) and radius r

center of circle (2) =(0,0) and radius 3

distance between their centers =(h0)2+(k0)2=(h)2+(k)2±r

So, r=r3

2r=3

r=32

from equation (2),

h2+k2=r2

(12)2+k2=(32)2

k2=9414 =84=2

k=±2

So, the center of circle (1) =(h,k)=(12,2)or(12,2)

Hence, option (d) is correct.

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