JEE MATHEMATICS MCQ: SYSTEM OF CIRCLES

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Application +4 | -1

The centre of the circle, which cuts orthogonally each of the three circles

{x^2} + {y^2} + 2x + 17y + 4 = 0

and

{x^2} + {y^2} + 7x + 6y + 11 = 0,{x^2} + {y^2} - x + 22y + 3 = 0

, is

View answer and explanation

Correct option: A

\left( {3,2} \right)

Option A (3,2) is correct because it represents the center of the circle that intersects the three given circles orthogonally, found by solving the equations 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