# Class 11 NCERT Solutions- Chapter 11 Conic Section – Exercise 11.1

**In each of the following Exercise 1 to 5, find the equation of the circle with**

**Question 1: Centre (0, 2) and radius 2**

**Solution:**

Given: Centre (0, 2) and radius (r) = 2

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

^{2}+ (y – k)^{2}= r^{2}As, Centre (h, k) = (0, 2) and radius (r) = 2

Thus, the equation of the circle is

(x – 0)

^{2}+ (y – 2)^{2}= 2^{2 }[by using formula (a– b)^{2}= a^{2}– 2ab + b^{2}]x

^{2}+ y^{2}+ 4 – 4y = 4x

^{2}+ y^{2}– 4y = 0Therefore, the equation of the circle is x

^{2}+ y^{2}– 4y = 0

**Question 2: Centre (–2, 3) and radius 4**

**Solution:**

Given: Centre (-2, 3) and radius (r) = 4

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

^{2}+ (y – k)^{2}= r^{2}As, centre (h, k) = (-2, 3) and radius (r) = 4

Thus, the equation of the circle is

(x + 2)

^{2}+ (y – 3)^{2}= (4)^{2 }[by using formula (a – b)^{2}= a^{2}– 2ab + b^{2}]x

^{2}+ 4x + 4 + y^{2}– 6y + 9 = 16x

^{2}+ y^{2}+ 4x – 6y – 3 = 0Therefore, the equation of the circle is x

^{2}+ y^{2}+ 4x – 6y – 3 = 0

**Question 3: Centre (1/2, 1/4) and radius (1/12)**

**Solution:**

Given: Centre (1/2, 1/4) and radius 1/12

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

^{2}+ (y – k)^{2}= r^{2}So, centre (h, k) = (1/2, 1/4) and radius (r) = 1/12

Thus, the equation of the circle is

(x – 1/2)

^{2}+ (y – 1/4)^{2}= (1/12)^{2 }[by using formula (a – b)^{2}= a^{2}– 2ab + b^{2}]x

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

^{2}– x + 1/4 + y^{2}– y/2 + 1/16 = 1/144144x

^{2}– 144x + 36 + 144y^{2}– 72y + 9 – 1 = 0144x

^{2}– 144x + 144y^{2}– 72y + 44 = 036x

^{2}+ 36x + 36y2 – 18y + 11 = 036x

^{2}+ 36y2 – 36x – 18y + 11= 0Therefore, the equation of the circle is 36x

^{2}+ 36y^{2}– 36x – 18y + 11= 0

**Question 4: Centre (1, 1) and radius √2**

**Solution:**

Given: Centre (1, 1) and radius √2

^{2}+ (y – k)^{2}= r^{2}So, centre (h, k) = (1, 1) and radius (r) = √2

Thus, the equation of the circle is

(x-1)

^{2}+ (y-1)^{2}= (√2)^{2 }[by using formula (a – b)^{2}= a^{2}– 2ab + b^{2}]x

^{2}– 2x + 1 + y^{2}-2y + 1 = 2x

^{2}+ y^{2}– 2x -2y = 0Therefore, the equation of the circle is x

^{2}+ y^{2}– 2x -2y = 0

**Question 5: Centre (–a, –b) and radius √(a**^{2} – b^{2})

^{2}– b

^{2})

**Solution:**

Given: Centre (-a, -b) and radius √(a

^{2}– b^{2})^{2}+ (y – k)^{2}= r^{2}So, centre (h, k) = (-a, -b) and radius (r) = √(a

^{2}– b^{2})Thus, the equation of the circle is

(x + a)

^{2}+ (y + b)^{2}= (√(a^{2}– b^{2})^{2}) [by using formula (a + b)^{2}= a^{2}+ 2ab + b^{2}]x

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

^{2}+ y^{2}+2ax + 2by + 2b^{2}= 0Therefore, the equation of the circle is x

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

**In each of the following Exercise 6 to 9, find the centre and radius of the circles.**

**Question 6: (x + 5)**^{2} + (y – 3)^{2} = 36

^{2}+ (y – 3)

^{2}= 36

**Solution:**

Given equation: (x + 5)

^{2}+ (y – 3)^{2}= 36(x – (-5))

^{2}+ (y – 3)^{2}= 6^{2}The equation is of the form (x – h)

^{2}+ (y – k)^{2}= r^{2 }where, h = -5, k = 3 and r = 6Therefore, the centre is (-5, 3) and its radius is 6.

**Question 7: x**^{2} + y^{2} – 4x – 8y – 45 = 0

^{2}+ y

^{2}– 4x – 8y – 45 = 0

**Solution:**

Given equation: 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) + 4^{2}) – 4 – 16 = 45(x – 2)

^{2}+ (y – 4)^{2 }= 65(x – 2)

^{2}+ (y – 4)^{2}= (√65)^{2}The equation is of the form (x-h)

^{2}+(y-k)^{2}= r^{2},where h = 2, k = 4 and r = √65Therefore, the centre is (2, 4) and its radius is √65.

**Question 8: x**^{2} + y^{2} – 8x + 10y – 12 = 0

^{2}+ y

^{2}– 8x + 10y – 12 = 0

**Solution:**

Given equation: 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}The equation is of the form (x-h)

^{2}+(y-k)^{2}= r^{2},where h = 4, k= -5 and r = √53Therefore, the centre is (4, -5) and its radius is √53.

**Question 9: 2x**^{2} + 2y^{2} – x = 0

^{2}+ 2y

^{2}– x = 0

**Solution:**

Given equation: 2x

^{2}+ 2y^{2}– x = 0.2x

^{2}+ 2y^{2}– x = 0(2x

^{2}– x) + 2y^{2}= 0(x

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

^{2}+ (y – 0)^{2}= (1/4)^{2}The equation is of the form (x-h)

^{2}+(y-k)^{2}= r^{2}, where, h = 1/4, k = 0, and r = 1/4Therefore, the centre is (1/4, 0) and its radius is 1/4.

**Question 10: 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:**

The equation of the circle is (x – h)

^{2}+ (y – k)^{2}= r^{2}As the circle passes through points (4,1) and (6,5)

So, When the circle passes through (4,1)

(4 – h)

^{2}+ (1 – k)^{2}= r^{2}……………..(1)When the circle passes through (6,5)

(6 – h)

^{2}+ (5 – k)^{2 }= r^{2}………………(2)Given that, the centre (h, k) of the circle lies on line 4x + y = 16,

4h + k =16 ………………… (3)

From equation (1) and (2), we get

(4 – h)

^{2}+ (1 – k)^{2}=(6 – h)^{2}+ (5 – k)^{2}16 – 8h + h

^{2}+1 -2k +k^{2}= 36 -12h +h^{2}+15 – 10k + k^{2}16 – 8h +1 -2k + 12h -25 -10k

4h +8k = 44

h + 2k = 11 ……………. (4)

Now let us multiply equation (3) by 2, and subtracting it with equation (4), we get

(h + 2k) – 2(4h + k) = 11 – 32

h + 2k – 8h – 2k = -21

-7h = -21

h = 3

Substitute this value of h in equation (4), we get

3 + 2k = 11

2k = 11 – 3

2k = 8

k = 4

We get h = 3 and k = 4

When we substitute the values of h and k in equation (1), we get

(4 – 3)

^{2}+ (1 – 4)^{2}= r^{2}(1)

^{2}+ (-3)^{2}= r^{2}1+9 = r

^{2}r = √10

Now, the equation of the circle is,

(x – 3)

^{2 }+ (y – 4)^{2}= (√10)^{2}x

^{2}– 6x + 9 + y^{2}– 8y + 16 =10x

^{2}+ y^{2 }– 6x – 8y + 15 = 0Therefore, the equation of the circle is x

^{2}+ y^{2}– 6x – 8y + 15 = 0

**Question 11: 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:**

The equation of the circle is (x – h)

^{2}+ (y – k)^{2}= r^{2}As the circle passes through points (2,3) and (-1,1)

So, When the circle passes through (2,3)

(2 – h)

^{2}+ (3 – k)^{2}=r^{2}……………..(1)When the circle passes through (-1,1)

(-1 – h)

^{2}+ (1– k)^{2}=r^{2}………………(2)Given that, the centre (h, k) of the circle lies on line x – 3y – 11= 0,

h – 3k =11 ………………… (3)

From the equation (1) and (2), we get

(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)

Now let us multiply equation (3) by 6, and subtracting it with equation 4, we get

6h+ 4k – 6(h-3k) = 11 – 66

6h + 4k – 6h + 18k = 11 – 66

22 k = – 55

k = -5/2

Substitute this value of k in equation (4), we get

6h + 4(-5/2) = 11

6h – 10 = 11

6h = 21

h = 21/6

h = 7/2

We get 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}Now, the equation of the circle is,

(x – 7/2)

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

^{2}+ [(2y+5)/2]^{2}= 130/44×2 -28x + 49 +4y

^{2}+ 20y + 25 =1304x

^{2}+4y^{2}-28x + 20y – 56 = 04(x

^{2}+y^{2}-7x + 5y – 14) = 0x

^{2}+ y^{2}– 7x + 5y – 14 = 0Therefore, the equation of the circle is x

^{2}+ y^{2}– 7x + 5y – 14 = 0

**Question 12: Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).**

**Solution:**

The equation of the circle is (x – h)

^{2}+ (y – k)^{2}= r^{2}Given the radius of the circle is 5 and it’s centre lies on the x-axis, k = 0 and r = 5.

So now, the equation of the circle is (x – h)

^{2}+ y^{2}= 25.Also given that the circle passes through the point (2, 3).

Therefore,

(2 – h)

^{2}+ 3^{2}= 25(2 – h)

^{2}= 25-9(2 – h)

^{2}= 162 – h = ± √16 = ± 4

If 2-h = 4, then h = -2

If 2-h = -4, then h = 6

Now, when h = -2, the equation of the circle is

(x + 2)

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

^{2}+ 4x + 4 + y^{2}= 25x

^{2}+ y^{2}+ 4x – 21 = 0Now, when h = 6, the equation of the circle is

(x – 6)

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

^{2}-12x + 36 + y^{2}= 25x

^{2}+ y^{2}-12x + 11 = 0Therefore, the equation of the circle is x

^{2}+ y^{2}– 4x + 21 = 0 and x^{2}+ y^{2}-12x + 11 = 0

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

**Solution:**

The equation of the circle is (x – h)

^{2}+ (y – k)^{2}= r^{2}When the circle passes through (0, 0),we get,

(0 – h)

^{2}+ (0 – k)^{2}= r^{2}h

^{2}+ k^{2}= r^{2}The equation of the circle is (x – h)

^{2}+ (y – k)^{2}= h^{2}+ k^{2}.Given that the circle intercepts points a and b on the coordinate axes.

Since, the circle passes through points (a, 0) and (0, b).

So the equations are,

(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 get

a

^{2}– 2ah + h^{2}+ k^{2}= h^{2}+ k^{2}a

^{2}– 2ah = 0a(a – 2h) =0

a = 0 or (a -2h) = 0

As, a ≠ 0; hence, (a -2h) = 0

h = a/2

From equation (2), we get

h

^{2}– 2bk + k^{2}+ b^{2 }= h^{2}+ k^{2}b

^{2}– 2bk = 0b(b– 2k) = 0

b= 0 or (b-2k) =0

As, a ≠ 0; hence, (b -2k) = 0

k = b/2

Now, substituting the value of h and k, we get

(x – a/2)

^{2}+ (y – b/4)^{2 }= (a/2)^{2 }+ (b/2)^{2}[(2x-a)/2]

^{2}+ [(2y+b)/2]^{2}= (a2 + b2)/44x

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

^{2}+ 4y^{2}-4ax – 4by = 04(x

^{2}+y^{2}-7x + 5y – 14) = 0x

^{2}+ y^{2}– ax – by = 0Therefore, the equation of the circle is x

^{2}+ y^{2}– ax – by = 0

**Question 14: Find the equation of a circle with centre (2,2) and passes through the point (4,5).**

**Solution:**

Given the centre of the circle as (h, k) = (2,2)

Also given that the circle passes through the 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

Now, the equation of the circle is

(x– h)

^{2}+ (y – k)^{2}= r^{2}(x –h)

^{2}+ (y – k)^{2}= (√13)^{2}(x –2)

^{2}+ (y – 2)^{2 }= (√13)^{2}x

^{2}– 4x + 4 + y^{2}– 4y + 4 = 13x

^{2}+ y^{2}– 4x – 4y = 5Therefore, the equation of the circle is x

^{2}+ y^{2}– 4x – 4y = 5

**Question 15: Does the point (–2.5, 3.5) lie inside, outside or on the circle x**^{2} + y^{2} = 25?

^{2}+ y

^{2}= 25?

**Solution:**

Given equation of the circle is x

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

^{2}+ y^{2}= 25(x – 0)

^{2}+ (y – 0)^{2}= 5^{2}The equation is of the form (x – h)

^{2}+ (y – k)^{2}= r^{2 },where, h = 0, k = 0 and r = 5.Now the distance between the point (-2.5, 3.5) and the centre (0,0) is

= √[(-2.5 – 0)

^{2}+ (-3.5 – 0)^{2}]= √(6.25 + 12.25)

= √18.5

= 4.3 [which is < 5]

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

Therefore, the point (-2.5, -3.5) lies inside the circle.