Learn Exercise 11.1 with Free Lessons & Tips

The equation of a circle with centre (h, k) and radius r is given as

(x­ – h)² + (y ­– k)² = r²

It is given that centre (h, k) = (0, 2) and radius (r) = 2.

Therefore, the equation of the circle is

(x – 0)² + (y – 2)² = 2²

x² + y² + 4 ­– 4 y = 4

x² + y² ­– 4y = 0

The equation of a circle with centre (h, k) and radius r is given as

(x­ – h)² + (y ­– k)² = r²

It is given that centre (h, k) = (–2, 3) and radius (r) = 4.

Therefore, the equation of the circle is

(x + 2)² + (y – 3)² = (4)²

x² + 4x + 4 + y² – 6y + 9 = 16

x² + y² + 4x – 6y – 3 = 0

The equation of a circle with centre (h, k) and radius r is given as

(x­ – h)² + (y ­– k)² = r²

It is given that centre (h, k) = and radius (r) =.

Therefore, the equation of the circle is

The equation of a circle with centre (h, k) and radius r is given as

(x­ – h)² + (y ­– k)² = r²

It is given that centre (h, k) = (1, 1) and radius (r) =.

Therefore, the equation of the circle is

The equation of a circle with centre (h, k) and radius r is given as

(x­ – h)² + (y ­– k)² = r²

It is given that centre (h, k) = (–a, –b) and radius (r) =.

Therefore, the equation of the circle is

The equation of the given circle is (x + 5)² + (y – 3)² = 36.

(x + 5)² + (y – 3)² = 36

⇒ {x – (–5)}² + (y – 3)² = 6², which is of the form (x – h)² + (y – k)² = r², where h = –5, k = 3, and r = 6.

Thus, the centre of the given circle is (–5, 3), while its radius is 6.

The equation of the given circle is x² + y² – 4x – 8y – 45 = 0.

x² + y² – 4x – 8y – 45 = 0

⇒ (x² – 4x) + (y² – 8y) = 45

⇒ {x² – 2(x)(2) + 2²} + {y² – 2(y)(4)+ 4²} – 4 –16 = 45

⇒ (x – 2)² + (y –4)² = 65

⇒ (x – 2)² + (y –4)² = , which is of the form (x – h)² + (y – k)² = r², where h = 2, k = 4, and .

Thus, the centre of the given circle is (2, 4), while its radius is.

The equation of the given circle is x² + y² – 8x + 10y – 12 = 0.

x² + y² – 8x + 10y – 12 = 0

⇒ (x² – 8x) + (y² + 10y) = 12

⇒ {x² – 2(x)(4) + 4²} + {y² + 2(y)(5) + 5²}– 16 – 25 = 12

⇒ (x – 4)² + (y + 5)² = 53

, which is of the form (x – h)² + (y – k)² = r², where h = 4, k = –5, and .

Thus, the centre of the given circle is (4, –5), while its radius is.

The equation of the given circle is 2x² + 2y² – x = 0.

, which is of the form (x – h)² + (y – k)² = r², where h = , k = 0, and .

Thus, the centre of the given circle is, while its radius is.

Let the equation of the required circle be (x – h)² + (y – k)² = r².

Since the circle passes through points (4, 1) and (6, 5),

(4 – h)² + (1 – k)² = r² … (1)

(6 – h)² + (5 – k)² = r² … (2)

Since the centre (h, k) of the circle lies on line 4x + y = 16,

4h + k = 16 … (3)

From equations (1) and (2), we obtain

(4 – h)² + (1 – k)² = (6 – h)² + (5 – k)²

⇒ 16 – 8h + h² + 1 – 2k + k² = 36 – 12h + h² + 25 – 10k + k²

⇒ 16 – 8h + 1 – 2k = 36 – 12h + 25 – 10k

⇒ 4h + 8k = 44

⇒ h + 2k = 11 … (4)

On solving equations (3) and (4), we obtain h = 3 and k = 4.

On substituting the values of h and k in equation (1), we obtain

(4 – 3)² + (1 – 4)² = r²

⇒ (1)² + (– 3)² = r²

⇒ 1 + 9 = r²

⇒ r² = 10

⇒

Thus, the equation of the required circle is

(x – 3)² + (y – 4)² =

x² – 6x + 9 + y² ­– 8y + 16 = 10

x² + y² – 6x – 8y + 15 = 0

Let the equation of the required circle be (x – h)² + (y – k)² = r².

Since the circle passes through points (2, 3) and (–1, 1),

(2 – h)² + (3 – k)² = r² … (1)

(–1 – h)² + (1 – k)² = r² … (2)

Since the centre (h, k) of the circle lies on line x – 3y – 11 = 0,

h – 3k = 11 … (3)

From equations (1) and (2), we obtain

(2 – h)² + (3 – k)² = (–1 – h)² + (1 – k)²

⇒ 4 – 4h + h² + 9 – 6k + k² = 1 + 2h + h² + 1 – 2k + k²

⇒ 4 – 4h + 9 – 6k = 1 + 2h + 1 – 2k

⇒ 6h + 4k = 11 … (4)

On solving equations (3) and (4), we obtain.

On substituting the values of h and k in equation (1), we obtain

Thus, the equation of the required circle is

Let the equation of the required circle be (x – h)² + (y – k)² = r².

Since the radius of the circle is 5 and its centre lies on the x-axis, k = 0 and r = 5.

Now, the equation of the circle becomes (x – h)² + y² = 25.

It is given that the circle passes through point (2, 3).

When h = –2, the equation of the circle becomes

(x + 2)² + y² = 25

x² + 4x + 4 + y² = 25

x² + y² + 4x – 21 = 0

When h = 6, the equation of the circle becomes

(x – 6)² + y² = 25

x² – 12x +36 + y² = 25

x² + y² – 12x + 11 = 0

Let the equation of the required circle be (x – h)² + (y – k)² = r².

Since the circle passes through (0, 0),

(0 – h)² + (0 – k)² = r²

⇒ h² + k² = r²

The equation of the circle now becomes (x – h)² + (y – k)² = h² + k².

It is given that the circle makes intercepts a and b on the coordinate axes. This means that the circle passes through points (a, 0) and (0, b). Therefore,

(a – h)² + (0 – k)² = h² + k² … (1)

(0 – h)² + (b – k)² = h² + k² … (2)

From equation (1), we obtain

a² – 2ah + h² + k² = h² + k²

⇒ a² – 2ah = 0

⇒ a(a – 2h) = 0

⇒ a = 0 or (a – 2h) = 0

However, a ≠ 0; hence, (a – 2h) = 0 ⇒ h =.

From equation (2), we obtain

h² + b² – 2bk + k² = h² + k²

⇒ b² – 2bk = 0

⇒ b(b – 2k) = 0

⇒ b = 0 or(b – 2k) = 0

However, b ≠ 0; hence, (b – 2k) = 0 ⇒ k =.

Thus, the equation of the required circle is

The centre of the circle is given as (h, k) = (2, 2).

Since the circle passes through point (4, 5), the radius (r) of the circle is the distance between the points (2, 2) and (4, 5).

Thus, the equation of the circle is