Home
Class 12
MATHS
The equationof circle passing through th...

The equationof circle passing through the intersection of `x^(2)+y^(2) = 4 `and `x^(2) +y^(2) -2x - 4y + 4=0` and touching the line `x +2y =0` is `:`

Text Solution

Verified by Experts

The correct Answer is:
A, B, D
Promotional Banner

Similar Questions

Explore conceptually related problems

Find the equation of the circle passing through the intersection of the circles x^2 + y^2-4 = 0 and x^2+y^2-2x-4y+4=0 and touching the line x + 2y=0

The locus of the centre of the circle passing through the intersection of the circles x^2+y^2= 1 and x^2 + y^2-2x+y=0 is

The angle of intersection of the circles x^(2)+y^(2)=4 and x^(2)+y^(2)+2x+2y , is

Find the equation of the circle passing through the point of intersection of the circles x^2 + y^2 - 6x + 2y + 4 = 0, x^2 + y^2 + 2x - 4y -6 = 0 and with its centre on the line y = x.

The equation of the circle passing through the point of intersection of the circles x^2+y^2-4x-2y=8 and x^2+y^2-2x-4y=8 and the point (-1,4) is (a) x^2+y^2+4x+4y-8=0 (b) x^2+y^2-3x+4y+8=0 (c) x^2+y^2+x+y=0 (d) x^2+y^2-3x-3y-8=0

The equation of the circle passing through the point of intersection of the circles x^2+y^2-4x-2y=8 and x^2+y^2-2x-4y=8 and the point (-1,4) is x^2+y^2+4x+4y-8=0 x^2+y^2-3x+4y+8=0 x^2+y^2+x+y=0 x^2+y^2-3x-3y-8=0

The equation of the circle passing through the point of intersection of the circles x^2+y^2-4x-2y=8 and x^2+y^2-2x-4y=8 and the point (-1,4) is x^2+y^2+4x+4y-8=0 x^2+y^2-3x+4y+8=0 x^2+y^2+x+y=0 x^2+y^2-3x-3y-8=0

Find the equation of the circle passing through the points of intersection of the circles x^2 + y^2 - 2x - 4y - 4 = 0 and x^2 + y^2 - 10x - 12y +40 = 0 and whose radius is 4.

The number of distinct straight lines through the points of intersection of x^(2) - y^(2) = 1" and " x^(2) + y^(2) - 4x - 5 = 0

Find the equation of the circle through points of intersection of the circle x^2 + y^2 -2x - 4y + 4 =0 and the line x + 2y = 4 which touches the line x + 2y = 0 .