**Find the equation of the circle with centre (0, 2) and radius 2**

Solution:

The equation of a circle with centre (h, k) and radius r is given as (x – h)^{2} + (y – k)^{2} = r^{2}

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} = 2^{2}

⟹ x^{2} + y^{2} + 4 – 4 y = 4

⟹ x^{2} + y^{2} – 4y = 0

**Find the equation of the circle with centre (–2, 3) and radius 4**

Solution:

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

(x – h)^{2} + (y – k)^{2} = r^{2}

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}

⟹ x^{2} + 4x + 4 + y^{2} – 6y + 9 = 16

⟹ x^{2} + y^{2} + 4x – 6y – 3 = 0

**Find the equation of the circle with centre(**^{1}/_{2},^{1}/_{4}) and radius(^{1}/_{12})

Solution:

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

(x – h)^{2} + (y – k)^{2} = r^{2}

It is given that centre (h, k) =(^{1}/_{2}, ^{1}/_{4}) and radius (r) = (^{1}/_{12})

Therefore, the equation of the circle is

(x – ^{1}/_{2})^{2} + (y – ^{1}/_{4})^{2} = (^{1}/_{12})^{2}

x^{2} – x + ^{1}/_{4} + y^{2} – ^{y}/_{2} + ^{1}/_{16} = ^{1}/_{144}

x^{2} – x + ^{1}/_{4} + y^{2} – ^{y}/_{2} + ^{1}/_{16} – ^{1}/_{144} = 0

144x^{2} – 144x + 36 + 144y^{2} – 72y + 44 = 0

36x^{2} – 36x + 36y^{2} – 18y + 11 = 0

36x^{2} + 36y^{2} – 36x – 18y + 11=0

**Find the equation of the circle with centre (1, 1) and radius √2.**

Solution:

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

(x – h)^{2} + (y – k)^{2} = r^{2}

It is given that centre (h, k) = (1, 1) and radius (r) = √2.

Therefore, the equation of the circle is

(x – 1)^{2} + (y – 1)^{2} = (√2)^{2}

x^{2} – 2x + 1+y^{2} – 2y + 1 = 2

x^{2} + y^{2} -2x – 2y =0

**Find the equation of the circle with centre (-a, -b) and radius √(a**^{2}– b^{2}).

Solution:

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

(x – h)^{2} + (y – k)^{2} = r^{2}

It is given that centre (h, k) = (-a, -b) and radius (r) = √(a^{2} – b^{2}).

Therefore, the equation of the circle is

(x + a)^{2} + (y + b)^{2} = (√(a^{2} – b^{2}))^{2}

x^{2} + 2ax + a^{2 }+y^{2} + 2by + b^{2} = a^{2} – b^{2}

x^{2} + y^{2} + 2ax + 2by + 2b^{2} = 0

**Find the centre and radius of the circle (x + 5)**^{2}+ (y – 3)^{2}= 36

Solution:

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

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

**Find the centre and radius of the circle x**^{2}+ y^{2}– 4x – 8y – 45 = 0

Solution:

The equation of the given circle is x^{2} + y^{2} – 4x – 8y – 45 = 0.

x^{2} + y^{2} – 4x – 8y – 45 = 0

⇒ (x^{2} – 4x) + (y^{2} – 8y) = 45

⇒ {x^{2} – 2(x)(2) + 2^{2}} + {y^{2} – 2(y)(4)+ 42} – 4 –16 = 45

⇒ (x – 2)^{2} + (y –4)^{2} = 65

⇒ (x – 2)^{2} + (y – 4)^{2}= 65 ,

Which is of the form (x – h)^{2} + (y – k)^{2} = r^{2}, where h = 2, k = 4 and since r^{2} = 65, we have r = √65 ,

Thus, the centre of the given circle is (2, 4), while its radius is √65 .

**Find the centre and radius of the circle x**^{2}+ y^{2}– 8x + 10y – 12 = 0

Solution:

The equation of the given circle is x^{2} + y^{2} – 8x + 10y – 12 = 0.

x^{2} + y^{2} – 8x + 10y – 12 = 0

⇒ (x^{2} – 8x) + (y^{2} + 10y) = 12

⇒ {x^{2} – 2(x)(4) + 4^{2}} + {y^{2} + 2(y)(5) + 5^{2}}– 16 – 25 = 12

⇒ (x – 4)^{2} + (y + 5)^{2} = 53

⟹ (x − 4)^{2} +{y − (−5)}^{2} = (√53)^{2} which is of the form (x – h)^{2} + (y – k)^{2} = r^{2}, where h = 4, k = – 5 and r = √53

Thus, the centre of the given circle is (4, –5), while its radius is √53.

**Find the centre and radius of the circle 2x**^{2}+ 2y^{2}– x = 0

Solution:

The equation of the given circle is 2x^{2} + 2y^{2} – x = 0.

2x^{2} + 2y^{2 } – x = 0

(2x^{2} – x)+2y^{2} = 0

2[(x^{2} – ^{x}/_{2})+y^{2}] = 0

{x^{2} – 2x(^{1}/_{4})+(^{1}/_{4})^{2} } + y^{2} – (^{1}/_{4})^{2} = 0

(x – ^{1}/_{4})^{2} + (y – 0)^{2} = (^{1}/_{4})^{2}

which is of the form (x – h)^{2} + (y – k)^{2} = r^{2}, where h = 1/4, k = 0 and r = 1/4.

Thus, the centre of the given circle is (1/4, 0), while its radius is 1/4.

**Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.**

Solution:

Let the equation of the required circle be (x – h)^{2} + (y – k)^{2} = r^{2}.

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

(4 – h)^{2} + (1 – k)^{2} = r^{2} …………………. (1)

(6 – h)^{2} + (5 – k)^{2} = r^{2} …………………. (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 + h^{2} + 1 – 2k + k^{2} = 36 – 12h + h^{2} + 25 – 10k + k^{2}

⇒ 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} = r^{2}

⇒ (1)^{2} + (– 3)^{2} = r^{2}

⇒ 1 + 9 = r^{2}

⇒ r^{2} = 10

⇒ r = √10

Thus, the equation of the required circle is

(x – 3)^{2} + ( y – 4)^{2} = (√10)^{2}

x^{2} – 6x + 9 + y^{2} – 8y + 16 = 10

x^{2} – 6x + y^{2} – 8y + 15 = 0

**Find the equation of the circle passing through the points (2, 3) and (–1, 1) and whose centre is on the line x – 3y – 11 = 0.**

Solution:

Let the equation of the required circle be (x – h)^{2} + (y – k)^{2} = r^{2}.

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

(2 – h)^{2} + (3 – k)^{2} = r^{2}——————- (1)

(–1 – h)^{2} + (1 – k)^{2} = r^{2} ————–(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 + h^{2} + 9 – 6k + k^{2} = 1 + 2h + h^{2} + 1 – 2k + k^{2}

⇒ 4 – 4h + 9 – 6k = 1 + 2h + 1 – 2k

⇒ 6h + 4k = 11 ————(4)

On solving equations (3) and (4), we obtain h =^{7}/_{2} and k =−^{5}/_{2}

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

(2 – ^{7}/_{2})^{2} + (3 + ^{5}/_{2})^{2} = r^{2}

(^{4-7}/_{2})^{2} + (^{6+5}/_{2})^{2} = r^{2}

(^{-3}/_{2})^{2} + (^{11}/_{2})^{2} = r^{2}

^{9}/_{4}+ ^{121}/_{4} = r^{2}

^{130}/_{4} = r^{2}

Thus the equation of the required circle is

(x – ^{7}/_{2})^{2} + (y + ^{5}/_{2})^{2} = ^{130}/_{4}

(^{2x-7}/_{2})^{2} + (^{2y+5}/_{2})^{2} = ^{130}/_{4}

4x^{2} – 28x + 49 + 4y^{2} + 20y + 25 = 130

4x^{2} + 4y^{2} – 28x + 20y – 56 = 0

4(x^{2} + y^{2} – 7x + 5y – 14) = 0

x^{2} + y^{2} – 7x + 5y – 14 = 0

**Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).**

Solution:

Let the equation of the required circle be (x – h)^{2} + (y – k)^{2} = r^{2}.

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} + y^{2} = 25.

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

( 2 – h)^{2} + 3^{2} = 25

( 2 – h)^{2} = 25 – 9 = 16

2 – h = ±√16 = ±4

If 2 – h = 4 then h = -2

If 2 – h = -4 then h = 6

when h = -2 the equation of the circle becomes

(x + 2)^{2} + y^{2} = 25

x^{2} + 4x + 4 + y^{2} – 25 = 0

x^{2} + 4x + y^{2} – 21 = 0

When h = 6 the equation of the circle becomes

(x – 6)^{2} + y^{2} = 25

x^{2} – 12x + 36 + y^{2} – 25 = 0

x^{2} – 12x + y^{2} + 11 = 0

**Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.**

solution:

Let the equation of the required circle be (x – h)^{2} + (y – k)^{2} = r^{2}.

Since the centre of the circle passes through (0, 0),

(0 – h)^{2} + (0 – k)^{2} = r^{2}

⇒ h^{2} + k^{2} = r^{2}

The equation of the circle now becomes (x – h)^{2} + ( y- k)^{2} =h^{2 }+ k^{2}

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} = h^{2} + k^{2}————- (1)

(0 – h)^{2} + (b – k)^{2} = h^{2} + k^{2}————— (2)

From equation (1), we obtain a^{2} – 2ah + h^{2} + k^{2} = h^{2} + k^{2}

⇒ a^{2} – 2ah = 0

⇒ a(a – 2h) = 0

⇒ a = 0 or (a – 2h) = 0

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

From equation (2), we obtain h^{2} + b^{2} – 2bk + k^{2} = h^{2} + k^{2}

⇒ b^{2} – 2bk = 0

⇒ b(b – 2k) = 0

⇒ b = 0 or(b – 2k) = 0

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

Thus, the equation of the required circle is

(x – ^{a}/_{2})^{2} + (y – ^{b}/_{2} )^{2} = ( ^{a}/_{2})^{2} +( ^{b}/_{2})^{2}

(^{2x-a}/_{2})^{2} + (^{2y-b}/_{2} )^{2} = [a^{2} +b^{2}]/_{2}

4x^{2} – 4ax +a^{2} + 4y^{2 }– 4by + b^{2} = a^{2} + b^{2}

4x^{2}+ 4y^{2}– 4ax– 4by = 0

x^{2}+ y^{2}– ax – by = 0

**Find the equation of a circle with centre (2, 2) and passes through the point (4, 5)**.

Solution:

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).

r = √[(2-4)^{2} + (2-5)^{2}] = √[(2)^{2} + (3)^{2}] = √(4+9) = √13

Thhus the equation of the circle is

( x- h)^{2} + ( y-k)^{2} = r^{2}

( x- 2)^{2} + ( y-2)^{2} = (√13)^{2}

x^{2} – 4x + 4 +y^{2} -4y + 4 = 13

x^{2} + y^{2}– 4x-4y-5 = 0

**Does the point (–2.5, 3.5) lie inside, outside or on the circle x**^{2}+ y^{2}= 25?

Solution:

The equation of the given circle is x^{2} + y^{2} = 25.

x^{2} + y^{2} = 25

⇒ (x – 0)^{2} + (y – 0)^{2} = 5^{2},which is of the form (x – h)^{2} + (y – k)^{2} = r^{2}, where h = 0, k = 0, and r = 5.

∴ Centre = (0, 0) and radius = 5

Distance between point (–2.5, 3.5) and centre (0, 0)

=√[(-2.5-0)^{2}+(3.5-0)^{2}]

=√(6.25+12.25)

=√18.5

≈4.3 <5

Since the distance between point (–2.5, 3.5) and centre (0, 0) of the circle is less than the radius of the circle, point (–2.5, 3.5) lies inside the circle.