## Question

### Solution

Correct option is

x2 + y2 ± 10x ± 8y + 16 = 0

âˆµ                       CM = NO = 4 ∴                      PC = 5

∴                      NC = 5

Centre of circle is (5, 4) ∴ Equation of circle, if centre in I quadrant

(x – 5)2 + (y – 4)2 = 25

If centre in II, III and IV quadrant, then equation are

(x + 5)2 + (y – 4)2 = 25,

(x + 5)2 + (y + 4)2 = 25

and              (x – 5)2 + (y + 4)2 = 25

Hence, there are 4 circles which satisfy the given conditions. They are

(x ± 5)2 + (y ± 4)2 = 25

or                x2 + y2 ± 10x ± 8y + 16 = 0.

#### SIMILAR QUESTIONS

Q1

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.

Q2

Find the equation if the circle the end points of whose diameter are the centres of the circle x2 + y2 + 6x – 14y = 1 and x2 + y2 – 4x + 10y = 2.

Q3

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.

Q4

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

Q5

The abscissa of two points A and B are the roots of the equation x2 + 2ax – b2 = 0 and their ordinates are the roots of the equation x2 + 2px –q2 = 0. Find the equation and the radius of the circle with AB as diameter.

Q6

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

Q7

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

Q8

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

Q9

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.

Q10

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.