Find the equation of a circle concentric with the circle 2x^2+2y^2+4x-6y-1=0 and touching the line 3x + 4y + 2 =0 ? Pls answer fast

Solution)
We have:2x2+2y2+4x-6y-1=0Dividing by 2 we get:x2+y2+2x-3y-12=0Center of this cirlce=-22,--32=-1,32As required cirlce is concentric with above cirlce hence centre of required circlewill also be -1,32Let equation of required cirlce be:x-h2+y-k2=r2, where h,k is centre and 'r' is radius.x+12+y-322=r2Cirlce touches 3x+4y+2=0 hence length of perpendicular from centre -1,32on 3x+4y+2=0 will give length of radius.r=-3+6+232+42=55=1Hence equation of cirlce is:x+12+y-322=1

  • 1
What are you looking for?