Home
Class 12
MATHS
Find the equation of the circle which cu...

Find the equation of the circle which cuts orthogonally the circle `x^2 + y^2 – 6x + 4y -3 = 0`,passes through (3,0) and touches the axis of y.

Promotional Banner

Similar Questions

Explore conceptually related problems

Find the equation of the circle which is concentric with the circle x^(2) + y^(2) - 6x - 4y- 3 = 0 , and has radius 5.

The equation of the circle concentric with the circle x^(2) + y^(2) - 6x - 4y - 12 =0 and touching y axis

Find the equation of the circle concentric with the circle x^2 +y^2 - 4x - 6y-9=0 and passing through the point (-4, -5)

Find the equation of the circle concentric with the circle x^(2) + y^(2) + 4x + 4y + 11 = 0 and passing through the point ( 5, 4).

Find the equation of the circle concentric With the circle x^(2) + y^(2) - 4x - 6y - 9 = 0 and passing through the point ( -4, -5).

Equation of the circle concentric with the circle x^(2) + y^(2) - 4x - 6y - 3 = 0 , and touching Y-axis , is

(i) Find the equation of a circle , which is concentric with the circle x^(2) + y^(2) - 6x + 12y + 15 = 0 and of double its radius. (ii) Find the equation of a circle , which is concentric with the circle x^(2) + y^(2) - 2x - 4y + 1 = 0 and whose radius is 5. (iii) Find the equation of the cricle concentric with x^(2) + y^(2) - 4x - 6y - 3 = 0 and which touches the y-axis. (iv) find the equation of a circle passing through the centre of the circle x^(2) + y^(2) + 8x + 10y - 7 = 0 and concentric with the circle 2x^(2) + 2y^(2) - 8x - 12y - 9 = 0 . (v) Find the equation of the circle concentric with the circle x^(2) + y^(2) + 4x - 8y - 6 = 0 and having radius double of its radius.

Find the equation of the circle , centred at (-1,4) , which touches the circle x^(2) + y^(2) - 6x - 2y + 1 = 0 externally

The loucs of the centre of the circle which cuts orthogonally the circle x^(2)+y^(2)-20x+4=0 and which touches x=2 is

Find the equation of the circle which cuts the circle x^(2)+y^(2)-14x-8y+64=0 and the coordinate axes orthogonally.