JEE MATHEMATICS MCQ: SYSTEM OF CIRCLES

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Application +4 | -1

The circle passing through the intersection of the circles,

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

and

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

having its centre on the line,

2x - 3y + 12 = 0,

also passes through the point

View answer and explanation

Correct option: D

(-3, 6)

Option D (-3, 6) is correct because it satisfies the equation of the circle formed by the intersection of the given circles and having its center on the given line.

More Questions from SYSTEM OF CIRCLES

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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

orthogonally, if The equation of the circle whose diameter is the common chord of the circles

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

and

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

, is If the circles

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

and

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

intersects orthogonally. Then, the value of

k

is If the circles

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

and

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

intersect in two distinct points

P

and

Q

, then the line

5 x+b y-a=0

passes through

P

and

Q

for The equation of the circle having its centre on the line

x + 2y - 3 = 0

and passing through the point of intersection of the circles

{x^2} + {y^2} - 4y + 1 = 0

and

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