The ends of a quadrant of a circle have the coordinates (1, 3) and (3, 1). Then the center of such a circle is

If the coordinate of one end of a diameter of a circle is (3,4) and the coordinates of its centre is (-3,2) then the coordinate of the other end of the diameter is .

The ends of diameter of a circle are (2,3) and (6,5) . The centre of the circle is

Line 2x+3y+1=0 is a tangent to the circle at (1,-1). This circle is orthogonal to a circle which is drawn having diameter as a line segment with end points (0, -1) and (-2, 3). Then the equation of the circle is

A circle touches the x-axis and also thouches the circle with center (0, 3) and radius 2. The locus of the center of the circle is (a)a circle (b) an ellipse (c)a parabola (d) a hyperbola

If one end of a diameter of a circle is (3,5) and the centre is (2,3) then the other end of the diameter is

Find the equation of the circle if the chord of the circle joining (1, 2) and (-3,1) subtents 90^0 at the center of the circle.

The line 2x-y+1=0 is tangent to the circle at the point (2, 5) and the center of the circle lies on x-2y=4. The radius of the circle is

The normal at the point (3, 4) on a circle cuts the circle at the point (-1,-2). Then the equation of the circle is

The line 2x-y+1=0 is tangent to the circle at the point (2, 5) and the center of the circle lies on x-2y=4 . Then find the radius of the circle.

The equation of the circle which touches the axes of coordinates and the line x/3+y/4=1 and whose center lies in the first quadrant is x^2+y^2-2c x-2c y+c^2=0 , where c is (a) 1 (b) 2 (c) 3 (d) 6