JEE MATHEMATICS MCQ: SYSTEM OF CIRCLES

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The locus of centres of family of circle passing through the origin and cutting the circle

{x^2} + {y^2} + 4x - 6y - 13 = 0

orthogonally, is

View answer and explanation

Correct option: D

4x - 6y - 13 = 0

Option D is correct because the locus of the centers is found by using the orthogonality condition of circles, resulting in the equation 4x - 6y - 13 = 0.

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x = 1

is the radical axis of the two orthogonally intersecting circles. If

{x^2} + {y^2} = 4

is one of the circles, then the other circle is The circle

{x^2} + {y^2} + 8y - 4 = 0

, cuts the real circle

{x^2} + {y^2} + gx + 4 = 0

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x^2+y^2+3 x+2 y+1=0

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x^2+y^2+3 x+4 y+2=0

, is If the circles

{x^2} + {y^2} + 2ky + 2x + 6 = 0

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{x^2} + {y^2} + 2ky + k = 0

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is If the circles

x^2+y^2+2 a x+c y+a=0

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x^2+y^2-3 a x+d y-1=0

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P

and

Q

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5 x+b y-a=0

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P

and

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x + 2y - 3 = 0

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