The equation circle passing through the point of intersection of x^2 + y^2 = 6 and x^2 + y^2 – 6x + 8 = 0
The equation circle passing through the point of intersection of x2 + y2 = 6 and x2 + y2 – 6x + 8 = 0 and the point (1, 1) is
(1) x2 – y2 – 3x + 1 = 0
(2) x2 – y2 + 3x + 1 = 0
(3) x2 + y2 – 3x + 1 = 0
(4) x2 + y2 + 3x + 1 = 0
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2 Answers
Option is (3)
Explanation
The equations of set of circles passing through the intersection of two given circle is
x^2+y^2-6x+8+k(x^2+y^2-6)=0
It also passes through (1,1)
So
1^1+1^2-6*1+8+k(1^1+1^2-6)=0
=>k =4/4=1
Hence the required equation
2x^2+2y^2-6x+8-6=0
=>x^2+y^2-3x+1=0
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