## Question

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.

easy

### Solution

Equation of circle having (–4, 3) and (12, –1) as the ends of a diameter is

(*x* + 4) (*x* – 12) + (*y* – 3) (*y* + 1) = 0

Comparing (1) with standard equation of circle

*x*^{2} + *y*^{2} + 2*gx* + 2*fy* + *c* = 0

Intercept on *y*-axis

#### SIMILAR QUESTIONS

Find the parametric form of the equation of the circle

*x*^{2} + *y*^{2} + *px* + *py* = 0

If the parametric of form of a circle is given by

(i) *x* = – 4 + 5 cos θ and *y* = – 3 + 5 sin θ

(ii) *x* = *a* cos α + b sin α and *y* = a sin α – b cos α

Find its Cartesian form.

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 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*.