Find the equation of the circle through the points of intersection of the circles `x² + y^2 - 4x – 6y - 12 = 0` and `x² + y^2 + 6x + 4y - 12 = 0` and cutting the circle `x² + y^2 - 2x - 4 = 0` orthogonally.
Topper's Solved these Questions
CIRCLE
FIITJEE|Exercise Solved Problems|34 Videos
CIRCLE
FIITJEE|Exercise Exercise 1|6 Videos
AREA
FIITJEE|Exercise Numerical Based|3 Videos
COMPLEX NUMBER
FIITJEE|Exercise NUMERICAL BASED|3 Videos
Similar Questions
Explore conceptually related problems
Find the equation of the circle through the points of interrection of the circles x^(2)+y^(2)-4x-6y-12=0 and x^(2)+y^(2)+6x+4y-12=0 and cutting the circle x^(2)+y^(2)-2x-4=0 orthogonally.
If the equation of the circle throught the points of intersection of the circles x^(2)+y^(2)-4x-6y-12=0 and x^(2)+y^(2)+"Ax"+"By"+C=0 and intersecting the circle x^(2)+y^(2)-2x-4=0 orthogonally is x^(2)+y^(2)+"Ax"+"By"+C=0 , then find teh value of (A+B+C) .
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 equation of the circle passing through the points of intersection of the circles x^(2)+y^(2)+6x+4y-12=0 , x^(2)+y^(2)-4x-6y-12=0 and having radius sqrt(13) is
The equation of the circle passing through the points of intersection of the circles x^(2)+y^(2)+6x+4y-12=0,x^(2)+y^(2)-4x-6y-12=0 and having radius sqrt(13) is
The equation of the circle passing through the points of intersection of the circles x^(2)+y^(2)+6x+4y-12=0 , x^(2)+y^(2)-4x-6y-12=0 and having radius sqrt(13) is
The equation of the circle passing through the points of intersection of the circles x^(2)+y^(2)+6x+4y-12=0,x^(2)+y^(2)-4x-6y-12=0 and having radius sqrt(13)" 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
