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The equation of the circle passing through the point of intersection ofthe circles `x^2+y^2-4x-2y=8`and `x^2+y^2-2x-4y=8`and the point `(-1,4)`is`x^2+y^2+4x+4y-8=0``x^2+y^2-3x+4y+8=0``x^2+y^2+x+y=0``x^2+y^2-3x-3y-8=0`A. `x^(2)+y^(2)+4x+4y-8=0`B. `x^(2)+y^(2)-3x+4y+8=0`C. `x^(2)+y^(2)+x+y-8=0`D. `x^(2)+y^(2)-3x-3y-8=0` |
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Answer» Correct Answer - 4 The equation of any circle through the points of intersection of the given circles is `x^(2)+y^(2)-4x-2y-8+k(x^(2)+y^(2)-2x-4y-8)=0` (1) Since circle (1) passes through `(-1,4)`, we have `k=1` Therefore, the requiredcircle is `x^(2)+y^(2)-3x-3y-8=0` |
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