The Centre of Circle: `(x+2)^(2)+(y+4)^2=16` is

Find the center of circle (x+2)^2+(y+4)^2=16

The centre and radius of the circle (x+ 2)^(2) + (y+4)^(2) = 9 are

Equation of the circle concentric with the circle x^(2) + y^(2) + 8x + 10 y - 7 = 0 , and passing through the centre of the circle x^(2) + y^(2) - 4x - 6y = 0 ,is

Find the equation of a circle which passes the centre of the circle x^(2)+y^(2)-6x =1 and concentric with the circle 2x^(2)+2y^(2)-8x +12y -1=0 .

Let the line segment joining the centres of the circles x^(2)-2x+y^(2)=0 and x^(2)+y^(2)+4x+8y+16=0 intersect the circles at P and Q respectively. Then the equation of the circle with PQ as its diameter is

Find the equation of the circle the end points of whose diameter are the centres of the circle : x^2 + y^2 + 6x - 14y=1 and x^2 + y^2 - 4x + 10y=2 .

Prove that the locus of the centre of the circle (1)/(2)(x^(2)+y^(2))+xcostheta+ysintheta-4=0 is x^(2)+y^(2)=1

Let A be the centre of the circle x^(2)+y^(2)-2x-4y-20=0 . If the tangents at the points B (1, 7) and D(4, -2) on the circle meet at the point C, then the perimeter of the quadrilateral ABCD is

Find the equation of the circle the end points of whose diameters are the centres of the circles x^2+y^2+16x-14y=1 and x^2+y^2-4x+10y=2

Prove that the centres of the three circles (x^2 +y^2 −4x−6y−12=0) , (x^2 +y^2 +2x+4y−5=0) and (x^2 +y^2 −10x−16y+7=0) are collinear.