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Question

# If the tangent to a circle x2+y2=5 at point (1,âˆ’2) touches the circle x2+y2âˆ’8x+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 circlex2+y2=5Equation of tangent of point (1,-2)xx1+yy1=5x(1)+y(−2)=5x−2y=5 - (1)Now this equation is a tangent to the other circle too.x2+y2−8x+6y+20=0Replace the term x from equation (1)(5+2y)2+y2−8(5+2y)+6y+20=0 On solving; 5y2+10y+5=0(y+1)2=0y=−1 So; x=5+2(−1)=3Hence the point of contact is (3,−1)

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