## Question

Find the centre and radius of the circle

2*x*^{2} + 2*y*^{2} = 3*x* – 5*y* + 7

### Solution

The given equation of circle is

2*x*^{2} + 2*y*^{2} = 3*x* –5*y* + 7

#### SIMILAR QUESTIONS

Find the equation of the image of the circle *x*^{2} + *y*^{2} + 16*x* – 24*y* + 183 = 0 by the line mirror 4*x* + 7*y* + 13 = 0.

Find the equation of the normal to the circle *x*^{2} + *y*^{2} = 2*x*, which is parallel to the line *x* + 2*y* = 3.

Find the equation of the circle which cuts orthogonally each of the three circles given below:

*x*^{2} + *y*^{2} – 2*x* + 3*y* – 7 = 0, *x*^{2} + *y*^{2} + 5*x* – 5*y* + 9 = 0 and *x*^{2} + *y*^{2} + 7*x* – 9*x* + 29 = 0.

Circum centre of the triangle PT_{1}T_{2} is at

If P is taken to be at (*h*, 0) such that P’ lies on the circle, the area of the rhombus is

Locus of mid-point of the chords of contact of *x*^{2} + *y*^{2} = 2 from the points on the line 3*x* + 4*y* = 10 is a circle with centre P. If O be the origin then OP is equal to

Suppose *ax* + *bx* + *c* = 0, where *a*, *b*, *c* are in A.P. be normal to a family or circles. The equation of the circle of the family which intersects the circle *x*^{2} + *y*^{2} – 4*x* – 4*y* – 1 = 0 orthogonally is

Find the equation of chord of *x*^{2} + *y*^{2} – 6*x* + 10*y* – 9 = 0 which is bisected at (–2, 4).

Find the equation of that chord of the *x*^{2} + *y*^{2} = 15 which is bisected at (3, 2).

Find the equation of the circle whose centre is the point of intersection of the lines 2*x* – 3*y* + 4 = 0 and 3*x* + 4*y* – 5 = 0 and passes through the origin.