**Question 1.**

**Question 2.**

**Question 3.Find the equation of the circle passing through the points (5, 7), (6, 6), and (2, -2).Solution:**

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

Since the required circle passes through points A(5, 7), B(6, 6), and D(2, -2),

CA = CB = CD = radius

Substituting k = 3 in (i), we get

h + 3(3) – 11 = 0

⇒ h + 9 – 11 = 0

⇒ h = 2

Centre of the circle is C(2, 3).

radius (r) = CD

**Question 4.Show that the points (3, -2), (1, 0), (-1, -2) and (1, -4) are concyclic.Solution:**

Let the equation of the circle passing through the points (3, -2), (1, 0) and (-1, -2) be

For point (3, -2),

Substituting x = 3 and y = -2 in (i), we get

9 + 4 + 6g – 4f + c = 0

⇒ 6g – 4f + c = -13 ….(ii)

For point (1, 0),

Substituting x = 1 andy = 0 in (i), we get

1 + 0 + 2g + 0 + c = 0

⇒ 2g + c = -1 ……(iii)

For point (-1, -2),

Substituting x = -1 and y = -2, we get

1 + 4 – 2g – 4f + c = 0

⇒ 2g + 4f – c = 5 …….(iv)

Adding (ii) and (iv), we get

8g = -8

⇒ g = -1

Substituting g = -1 in (iii), we get

-2 + c = -1

⇒ c = 1

Substituting g = -1 and c = 1 in (iv), we get

-2 + 4f – 1 = 5

⇒ 4f = 8

⇒ f = 2

= 1 + 16 – 2 – 16 + 1

= 0

= R.H.S.

Point (1, -4) satisfies equation (v).

∴ The given points are concyclic.