JEE MATHEMATICS MCQ: SYSTEM OF CIRCLES

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Application +4 | -1

Which of the following is a point on the common chord of the circles

{x^2} + {y^2} + 2x - 3y + 6 = 0

and

{x^2} + {y^2} + x - 8y - 13 = 0

View answer and explanation

Correct option: D

\left( {1,\; - 4} \right)

Option D

\left( {1,\; - 4} \right)

is correct because it is the only point that satisfies the equation of the common chord, which is found by subtracting the equations of the two circles.

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