**Question 1.Find the equation of a circle with(i) centre at origin and radius 4.(ii) centre at (-3, -2) and radius 6.(iii) centre at (2, -3) and radius 5.(iv) centre at (-3, -3) passing through point (-3, -6).Solution:**

(i) The equation of a circle with centre at origin and radius ‘r’ is given by

(iv) Centre of the circle is C (-3, -3) and it passes through the point P (-3, -6).

**Question 2.**

(iii) Given the equation of the circle is

**Question 3.Find the equation of the circle with centre(i) at (a, b) and touching the Y-axis.(ii) at (-2, 3) and touching the X-axis.(iii) on the X-axis and passing through the origin having radius 4.(iv) at (3, 1) and touching the line 8x – 15y + 25 = 0.Solution:**

(i) Since the circle is touching the Y-axis, the radius of the circle is X-co-ordinate of the centre.

(ii) Since the circle is touching the X-axis, the radius of the circle is the Y co-ordinate of the centre.

(iii) Let the co-ordinates of the centre of the required circle be C (h, 0).

Since the circle passes through the origin i.e., O(0, 0)

OC = radius

⇒ = 16

⇒ h = ±4

(iv) Centre of the circle is C (3, 1).

Let the circle touch the line 8x – 15y + 25 = 0 at point M.

CM = radius (r)

CM = Length of perpendicular from centre C(3, 1) on the line 8x – 15y + 25 = 0

**Question 4.Find the equation of the circle, if the equations of two diameters are 2x + y = 6 and 3x + 2y = 4 and radius is 9.Solution:**

Given equations of diameters are 2x + y = 6 and 3x + 2y = 4.

Let C (h, k) be the centre of the required circle.

Since point of intersection of diameters is the centre of the circle,

x = h, y = k

Equations of diameters become

2h + k = 6 …..(i)

and 3h + 2k = 4 ……..(ii)

By (ii) – 2 × (i), we get

-h = -8

⇒ h = 8

Substituting h = 8 in (i), we get

2(8) + k = 6

⇒ k = 6 – 16

⇒ k = -10

Centre of the circle is C (8, -10) and radius, r = 9

**Question 5.If y = 2x is a chord of the circle – 10x = 0, find the equation of the circle with this chord as diameter.Solution:**

**Question 6.Find the equation of a circle with a radius of 4 units and touch both the co-ordinate axes having centre in the third quadrant.**

**Solution:**

The radius of the circle = 4 units

Since the circle touches both the co-ordinate axes and its centre is in the third quadrant,

the centre of the circle is C(-4, -4).

**Question 7.Find the equation of the circle passing through the origin and having intercepts 4 and -5 on the co-ordinate axes.Solution:**

Let the circle intersect X-axis at point A and intersect Y-axis at point B.

the co-ordinates of point A are (4, 0) and the co-ordinates of point B are (0, -5).

**Question 8.Find the equation of a circle passing through the points (1, -4), (5, 2) and having its centre on line x – 2y + 9 = 0.Solution:**

Let C(h, k) be the centre of the required circle which lies on the line x – 2y + 9 = 0.

Equation of line becomes

h – 2k + 9 = 0 …..(i)

Also, the required circle passes through points A(1, -4) and B(5, 2).

CA = CB = radius

CA = CB

Substituting k = 3 in (i), we get

h – 2(3) + 9 = 0

⇒ h – 6 + 9 = 0

⇒ h = -3

Centre of the circle is C(-3, 3).

radius (r) = CA