**Q.1: A circle is given with centre (0, 3) and radius 3. Obtain the equation of the given circle.**

** **

**Sol:**

**The General equation of a circle with centre (h, k) and radius r is:**

**(a – h) ^{2} + (b – k)^{2} = r^{2}**

**Given:**

**(h, k) = centre of a circle = (0, 3) and radius r = 3**

**The equation of a circle is:**

(a – 0)^{2} + (b – 3)^{2} = 3^{2}

a^{2 }+ b^{2} – 6b + 9 = 9

a^{2 }+ b^{2} – 6b + 9 – 9 = 0

**a ^{2 }+ b^{2} – 6b = 0**

** **

**Q.2: A circle is given with centre (-1, 2) and radius 4. Find the equation of the circle.**

** **

**Sol:**

**The General equation of a circle with centre (h, k) and radius r is:**

**(a – h) ^{2} + (b – k)^{2} = r^{2}**

**Given:**

**Centre of a circle = (h, k) = (-1, 2) and radius (r) = 4**

**The equation of a circle is:**

(a – (- 1))^{2} + (b – 2)^{2} = 4^{2}

(a + 1)^{2} + (b – 2)^{2} = 4^{2}

a^{2 }+ 2a + 1 + b^{2} – 4b + 4 = 16

a^{2 }+ 2a + b^{2} – 4b + 5 = 16

a^{2 }+ 2a + b^{2} – 4b + 5 – 16 = 0

**a ^{2 }+ 2a + b^{2} – 4b – 11 = 0**

** **

**Q.3: A circle is given with centre ( 13, **

12 ) and radius 114 . Find the equation of the circle.

** **

**Sol:**

**The General equation of a circle with centre (h, k) and radius r is:**

**(a – h) ^{2} + (b – k)^{2} = r^{2}**

**Given:**

Centre of a circle = (h, k) = (

**The equation of a circle is:**

(a – ^{2} + (b – ^{2} = ^{2}

(a – ^{2} + (b – ^{2} = ^{2}

**Q.4: A circle is given with centre (2, 2) and radius 5–√. Obtain the equation of the given circle.**

** **

**Sol:**

**The General equation of a circle with centre (h, k) and radius r is:**

**(a – h) ^{2} + (b – k)^{2} = r^{2}**

Given:

Centre of a circle = (h, k) = (2, 2) and radius (r) =

**The equation of a circle is:**

(a – 2)^{2} + (b – 2)^{2} =

(a – 2)^{2} + (b – 2)^{2} =

a^{2 }– 4a + 4 + b^{2} – 4b + 4 = 5

a^{2 }– 4a + b^{2} – 4b + 8 = 5

a^{2 }– 4a + b^{2} – 4b + 8 – 5 = 0

a^{2 }– 4a + b^{2} – 4b + 3 = 0

**a ^{2 }+ b^{2} – 4a – 4b + 3 = 0**

** **

**Q.5: A circle is given with centre (- x, – y) and radius x2–y2−−−−−√. Find the equation of the circle.**

** **

**Sol:**

**The General equation of a circle with centre (h, k) and radius r is:**

**(a – h) ^{2} + (b – k)^{2} = r^{2}**

Given:

Centre of a circle = (h, k) = (- x, – y) and radius ‘r’ =

**The equation of a circle is:**

(a – (- x)^{2} + (b – (- y)^{2} =

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

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

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

**a ^{2 }+ 2ax + b^{2} + 2by + y^{2} = 0**

** **

**Q.6: The equation of a given circle is given as (a + 5) ^{2} + (b – 3)^{2} = 36. Find the centre and radius of the circle.**

** **

**Sol:**

Given:

The equation of a given circle (a + 5)^{2} + (b – 3)^{2} = 36

(a + 5)^{2} + (b – 3)^{2} = 36

{a – (- 5)}^{2} + (b – 3) = 6^{2}

Which is of the form (a – h) ^{2 }+ (b – k) ^{2} = r^{2}, where h = – 5, k = 3 and r = 6.

**Hence, the centre of the circle is (-5, 3), and the radius of circle is 6.**

**Q.7: The equation of a given circle is given as a ^{2 }+ b^{2} – 4a – 4b + 3 = 0**

**Find the radius and centre of the circle.**

** **

**Sol:**

Given:

The equation of a given circle a^{2 }+ b^{2} – 4a – 4b + 3 = 0

a^{2 }+ b^{2} – 4a – 4b + 3 = 0

a^{2 }– 4a + b^{2} – 4b + 3 = 0

{a^{2 }– 2.a.2 + 2^{2}} + {b^{2 }– 2.b.2 + 2^{2}} – 2^{2} – 2^{2} + 3 = 0

{a^{2 }– 2.a.2 + 2^{2}} + {b^{2 }– 2.b.2 + 2^{2}} – 8 + 3 = 0

{a^{2 }– 2.a.2 + 2^{2}} + {b^{2 }– 2.b.2 + 2^{2}} = 5

(a – 2)^{2} + (b – 2)^{2} =

Which is of the form (a – h) ^{2 }+ (b – k) ^{2} = r^{2}, where h = 2, k = 2 and r =

**Thus, the centre of the circle is (2, 2), and the radius of circle is 5–√.**

** **

**Q.8: The equation of a given circle is given as a ^{2 }+ b^{2} – 6a + 8b – 14 = 0**

**Find the centre and radius of the circle.**

** **

**Sol:**

**Given:**

** The equation of a given circle a ^{2 }+ b^{2} – 6a + 8b – 14 = 0**

a^{2 }– 6a + b^{2} + 8b – 14 = 0

{a^{2 }– 2.a.3 + 3^{2}} + {b^{2} + 2.b.4 + 4^{2}} – 3^{2 }– 4^{2 }– 14 = 0

(a – 3) ^{2} + (b + 4) ^{2} – 9 – 16 – 14 = 0

(a – 3) ^{2} + (b + 4) ^{2} – 39 = 0

(a – 3) ^{2} + (b + 4) ^{2} = 39

(a – 3) ^{2} + (b – (- 4)) ^{2} = 39

Which is of the form (a – h) ^{2 }+ (b – k) ^{2} = r^{2}, where h = 3, k = 4 and r =

**Thus, the centre of the circle is (3, 4), and the radius of circle is 39−−√.**

** **

**Q.9: The equation of a given circle is given as 2a ^{2 }+ 2b^{2} – 8a = 0**

**Find the centre and radius of the circle.**

** **

**Sol:**

Given:

The equation of a given circle 2a^{2 }+ 2b^{2} – 8a = 0

2a^{2 }+ 2b^{2} – 8a = 0

2a^{2 }– 8a + 2b^{2} = 0

2[a^{2} – 4a + b^{2}] = 0

{a^{2} – 2.a.2 + 2^{2}} – 2^{2} + (b – 0)^{2} = 0

[a^{2} – 2.a.2 + 2^{2}] + [b – 0]^{2} – 2^{2} = 0

(a – 2)^{2} + (b – 0)^{2} = 2^{2} = 4

Which is of the form (a – h) ^{2 }+ (b – k) ^{2} = r^{2}, where h = 2, k = 0 and r =

**Thus, the centre of the circle is (3, 4), and the radius of circle is 2.**

** **

**Q.10: The circle passing through the points (6, 2) and (4, 3) and whose centre lies on the line 2a + y = 8. Find the equation of the circle.**

** **

**Sol:**

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

**(a – h) ^{2} + (b – k)^{2} = r^{2}**

As the circle passes through (6, 2) and (4, 3)

(6 – h)^{2} + (2 – k)^{2} = r^{2} **. . . . . . . . . . . . . . . . (1)**

(4 – h)^{2} + (3 – k)^{2} = r^{2} **. . . . . . . . . . . . . . . . (2)**

The centre (h, k) of the circle lies on the line 2a + y = 8,

2h + k = 8 **. . . . . . . . . . . . . . . . . . . . . (3)**

**Now, On Comparing equations 1 and 2, we will get:**

(6 – h)^{2} + (2 – k)^{2} = (4 – h)^{2} + (3 – k)^{2}

36 – 12h + h^{2} + 4 – 4k + k^{2} = 16 – 8h + h^{2} + 9 – 6k + k^{2}

36 – 12h + 4 – 4k = 16 – 8h + 9 – 6k

– 12h – 4k + 8h + 6k = 16 + 9 – 36 – 4

4h – 2k = 15 **. . . . . . . . . . . . . . . . . . (4)**

**Now, on Solving Equations (3) and (4), we will get:**

h =

k =

**Substituting the values of h and k in equation (1), we will get:**

(6 – ^{2} + (2 – ^{2} = r^{2}

**The required equation is:**

(a – ^{2} + (b – ^{2} = r^{2}

(a – ^{2} + (b – ^{2} = 7.57^{2}

**Therefore, the required equation is: 64a ^{2}– 496a + 64b^{2} -32b = 2702.51**

**Q.11: The circle passing through the points (3, 2) and (-2, 2) and whose centre lies on the line a – 3y – 13 = 0. Find the equation of the circle.**

** **

**Sol:**

**The General equation of a circle with centre (h, k) and radius r is:**

**(a – h) ^{2} + (b – k)^{2} = r^{2}**

As the circle passes through (3, 2) and (-2, 1)

**(3 – h) ^{2} + (2 – k)^{2} = r^{2} . . . . . . . . . . . . . . . . (1)**

**(-2 – h) ^{2} + (2 – k)^{2} = r^{2} . . . . . . . . . . . . . . . (2)**

The centre (h, k) of the circle lies on the line a – 3y – 13 = 0,

**h – 3k = 13 . . . . . . . . . . . . . . . (3)**

**Now, on comparing equations 1 and 2, we will get:**

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

9 – 6h + h^{2} = 4 + 4h + h^{2}

9 – 6h = 4 + 4h

**h = 0.5 . . . . . . . . . . . . . . . . . . (4)**

**Now, on solving equation (3), we will get:**

k = –

**Now, on substituting the values of h and k in equation (1), we will get:**

(3 – 0.5)^{2} + (2 – (- ^{2} = r^{2}

(2.5) ^{2} + (2 + ^{2} = r^{2}

(2.5) ^{2} + (2 + 4.16)^{2} = r^{2}

(2.5) ^{2} + (6.16)^{2} = r^{2}

r^{2 }= 44.1956

r = 6.65

**The required equation is:**

(a – h)^{2} + (b – k)^{2} = r^{2}

(a – 0.5)^{2} + (b – (- 4.16))^{2} = r^{2}

(a – 0.5)^{2} + (b + 4.16)^{2} = r^{2}

a^{2} – a + 6.25 + b^{2} + 8.32b + 17.30 = 44.19

a^{2} – a + b^{2} + 8.32b = 44.19 – 6.25 – 17.30

**a ^{2} + b^{2} – a + 8.32b – 20.64 = 0**

** **

**Q.12: The radius of the circle is 6 whose centre lies on x – axis and passes through the point (3, 2). Find the equation.**

** **

**Sol:**

Given:

The radius of the circle is 6 and passes through the point (3, 2)

**The General equation of a circle with centre (h, k) and radius r is:**

**(a – h) ^{2} + (b – k)^{2} = r^{2}**

As, the centre lies on x – axis, k = 0, the equation becomes:

(a – h)^{ 2} + b^{ 2} = 36

(3 – h)^{ 2} + 2^{2} = 36

(3 – h)^{ 2} = 36 – 4

3 – h =

h =

**Q.13: ****The circle passing through (0, 0) and making intercepts a and b on the coordinate axes. Find the equation of the circle.**

** **

**Sol:**

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

The centre of the circle passes through (0, 0):

(0 – h) ^{2} + (0 – k) ^{2 }= r ^{2}

h ^{2} + k ^{2} = r ^{2}

Therefore, the equation of the circle is:

(x – h) ^{2} + (y – k) ^{2} = h ^{2} + k ^{2} .

Given, the circle makes intercepts a and b on the coordinate axes.

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)**

Now, 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

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

Now, from equation (2), we will get:

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;

Therefore, (b – 2k) = 0 ⇒ **k =b/2.**

**Thus, the equation of the required circle is:**

**Q.14: ****The circle with centre (3, 3) and passes through the point (6, 5), find the equation of a circle?**

** **

**Sol:**

**Given:**

**The centre of the circle (h, k) = (3, 3). **

As the circle passes through point (6, 5), the radius (r) of the circle is the distance between the points (2, 2) and (6, 5).

**Therefore, the equation of the circle is**:

**Q.****15: Check whether the point (–2.0, 3.0) lies inside, outside or on the circle x ^{2} + y ^{2} = 25? **

** **

**Sol:**

**Given:**

**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 (a – h)^{ 2} + (b – k)^{ 2} = r^{2}, where h = 0, k = 0, and r = 5.

Centre = (0, 0) and radius = 5

Distance between point (– 2.0, 3.0) and centre (0, 0)

=

As, the distance between point (– 2.0, 3.0) and centre (0, 0) of the circle is less than the radius of the circle, **point (– 2.0, 3.0) lies inside the circle.**