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Question

Let a circle passing through (4,0) touches the circle x2+y2+4x−6y−12=0, then radius of circle C is:

A
52
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B
57
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C
5
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D
4
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Solution

The correct option is C 5
Given,

x2+y2+4x6y12=0

Equation of tangent at (1,1)

xy+2(x+1)3(y1)12=0

3x4y7=0

Therefore the equation of circle is

(x2+y2+4x6y12)+λ(3x4y7)=0

It passes through (4,0) :

(16+1612)+λ(127)=0

20+λ(5)=0

λ=4

(x2+y2+4x6y12)4(3x4y7)=0

or x2+y28x+10y+16=0

Radius =16+2516=5

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