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Question

If the tangent to a circle x2+y2=5 at point (1,2) touches the circle x2+y28x+6y+20=0, then its point of contact is?

A
(2,1)
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B
(3,1)
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C
(1,3)
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D
(5,0)
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Solution

The correct option is B (3,1)
Equation of first circle

x2+y2=5

Equation of tangent of point (1,-2)
xx1+yy1=5
x(1)+y(2)=5
x2y=5 - (1)

Now this equation is a tangent to the other circle too.

x2+y28x+6y+20=0

Replace the term x from equation (1)

(5+2y)2+y28(5+2y)+6y+20=0
On solving; 5y2+10y+5=0


(y+1)2=0


y=1
So; x=5+2(1)=3

Hence the point of contact is (3,1)

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