## Question

Find the equation of the circle which passes through the origin and makes intercepts of length *a* and *b* on the *x* and *y* axes respectively.

### Solution

*x*^{2} + *y*^{2} – 8*x* – 10 *y* + 39 = 0

Let the equation of the circle be

*x*^{2} + *y*^{2} + 2*gx* + 2*fy* + *c* = 0 …(1)

since the circle passes through the origin, we get *c* = 0 and given the intercepts on *x* and *y* axes are *a* and *b*

∴

∴

Hence the equation of circle from (1) becomes

#### SIMILAR QUESTIONS

Find the equation if the circle the end points of whose diameter are the centres of the circle *x*^{2} + *y*^{2} + 6*x* – 14*y* = 1 and *x*^{2} + *y*^{2} – 4*x* + 10*y* = 2.

The sides of a square are *x* = 2, *x* = 3, *y* = 1and *y* = 2. Find the equation of the circle drawn on the diagonals of the square as its diameter.

Find the equation of the circum circle of the quadrilateral formed by the four lines *ax* + *by* ± *c* = 0 and *bx* – *ay* ± *c* = 0.

The abscissa of two points *A* and *B* are the roots of the equation *x*^{2} + 2*ax* – *b*^{2} = 0 and their ordinates are the roots of the equation *x*^{2} + 2*px* –*q*^{2} = 0. Find the equation and the radius of the circle with *AB* as diameter.

Find the equation of the circle which passes through the points (4, 1), (6, 5) and has its centre on the line 4*x* + *y* = 16.

Find the equation of the circle passing through the three non-collinear points (1, 1), (2, –1) and (3, 2).

Show that the four points (1, 0), (2, –7), (8, 1) and (9, –6) are concyclic.

Find the equation of the circle whose diameter is the line joining the points (–4, 3) and (12, –1). Find also the intercept made by it on *y*-axis.

Find the equation of the circle which touches the axis of *y* at a distance of 4 units from the origin and cuts the intercept of 6 units from the axis of *x*.

Find the equation of the circle which touches the axes and whose centre lies on the line *x* – 2*y* = 3.