JEE MATHEMATICS MCQ: SYSTEM OF CIRCLES

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Application +4 | -1

The locus of the centre of the circle which cuts the circles

x^2+y^2+4 x-6 y+9=0

and

x^2+y^2-4 x+

6 y+4=0

orthogonally is

View answer and explanation

Correct option: B

8 x-12 y+5=0

Option B (

8 x-12 y+5=0

) is correct because it represents the locus of the center of the circle that cuts the given circles orthogonally, derived from the orthogonality condition.

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