Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 6 Circle Ex 6.1 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 6 Circle Ex 6.1

Question 1.

Find the equation of a circle with

(i) centre at origin and radius 4.

(ii) centre at (-3, -2) and radius 6.

(iii) centre at (2, -3) and radius 5.

(iv) centre at (-3, -3) passing through point (-3, -6).

Solution:

(i) The equation of a circle with centre at origin and radius ‘r’ is given by

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

Here, r = 4

∴ The required equation of the circle is x^{2} + y^{2} = 4^{2} i.e., x^{2} + y^{2} = 16.

(ii) The equation of a circle with centre at (h, k) and radius ‘r’ is given by

(x – h)^{2} + (y – k)^{2} = r^{2}

Here, h = -3, k = -2 and r = 6

∴ The required equation of the circle is

[x – (-3)]^{2} + [y – (-2)]^{2} = 6^{2}

⇒ (x + 3)^{2} + (y + 2)^{2} = 36

⇒ x^{2} + 6x + 9 + y^{2} + 4y + 4 – 36 = 0

⇒ x^{2} + y^{2} + 6x + 4y – 23 = 0

(iii) The equation of a circle with centre at (h, k) and radius ‘r’ is given by

(x – h)^{2} + (y – k)^{2} = r^{2}

Here, h = 2, k = -3 and r = 5

The required equation of the circle is

(x – 2)^{2} + [y – (-3)]^{2} = 5^{2}

⇒ (x – 2)^{2} + (y + 3)^{2} = 25

⇒ x^{2} – 4x + 4 + y^{2} + 6y + 9 – 25 = 0

⇒ x^{2} + y^{2} – 4x + 6y – 12 = 0

(iv) Centre of the circle is C (-3, -3) and it passes through the point P (-3, -6).

The equation of a circle with centre at (h, k) and radius ‘r’ is given by

(x – h)^{2} + (y – k)^{2} = r^{2}

Here, h = -3, k = -3, r = 3

The required equation of the circle is

[x – (-3)]^{2} + [y – (-3)]^{2} = 3^{2}

⇒ (x + 3)^{2} + (y + 3)^{2} = 9

⇒ x^{2} + 6x + 9 + y^{2} + 6y + 9 – 9 = 0

⇒ x^{2} + y^{2} + 6x + 6y + 9 = 0

Check:

If the point (-3, -6) satisfies x^{2} + y^{2} + 6x + 6y + 9 = 0, then our answer is correct.

L.H.S. = x^{2} + y^{2} + 6x + 6y + 9

= (-3)^{2} + (-6)^{2} + 6(-3) – 6(-6) + 9

= 9 + 36 – 18 – 36 + 9

= 0

= R.H.S.

Thus, our answer is correct.

Question 2.

Find the centre and radius of the following circles:

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

(ii) (x – 5)^{2} + (y – 3)^{2} = 20

(iii) \(\left(x-\frac{1}{2}\right)^{2}+\left(y+\frac{1}{3}\right)^{2}=\frac{1}{36}\)

Solution:

(i) Given equation of the circle is

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

⇒ x^{2} + y^{2} = (5)^{2}

Comparing this equation with x^{2} + y^{2} = r^{2}, we get r = 5

Centre of the circle is (0, 0) and radius of the circle is 5.

(ii) Given equation of the circle is

(x – 5)^{2} + (y – 3)^{2} = 20

⇒ (x – 5)^{2} + (y – 3)^{2} = (√20)^{2}

Comparing this equation with (x – h)^{2} + (y – k)^{2} = r^{2}, we get

h = 5, k = 3 and r = √20 = 2√5

Centre of the circle = (h, k) = (5, 3)

and radius of the circle = 2√5.

(iii) Given the equation of the circle is

Comparing this equation with (x – h)^{2} + (y – k)^{2} = r^{2}, we get

h = \(\frac{1}{2}\), k = \(\frac{-1}{3}\) and r = \(\frac{1}{6}\)

Centre of the circle = (h, k) = (\(\frac{1}{2}\), \(\frac{-1}{3}\)) and radius of the circle = \(\frac{1}{6}\)

Question 3.

Find the equation of the circle with centre

(i) at (a, b) and touching the Y-axis.

(ii) at (-2, 3) and touching the X-axis.

(iii) on the X-axis and passing through the origin having radius 4.

(iv) at (3, 1) and touching the line 8x – 15y + 25 = 0.

Solution:

(i) Since the circle is touching the Y-axis, the radius of the circle is X-co-ordinate of the centre.

∴ r = a

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

(x – h)^{2} + (y – k)^{2} = r^{2}

Here, h = a, k = b

The required equation of the circle is

⇒ (x – a)^{2} + (y – b)^{2} = a^{2}

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

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

(ii) Since the circle is touching the X-axis, the radius of the circle is the Y co-ordinate of the centre.

∴ r = 3

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

(x – h)^{2} + (y – k)^{2} = r^{2}

Here, h = -2, k = 3

The required equation of the circle is

⇒ (x + 2)^{2} + (y – 3)^{2} = 3^{2}

⇒ x^{2} + 4x + 4 + y^{2} – 6y + 9 = 9

⇒ x^{2} + y^{2} + 4x – 6y + 4 = 0

(iii) Let the co-ordinates of the centre of the required circle be C (h, 0).

Since the circle passes through the origin i.e., O(0, 0)

OC = radius

⇒ \(\sqrt{(h-0)^{2}+(0-0)^{2}}=4\)

⇒ h^{2} = 16

⇒ h = ±4

the co-ordinates of the centre are (4, 0) or (-4, 0).

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

(x – h)^{2} + (y – k)^{2} = r^{2}

Here, h = ± 4, k = 0, r = 4

The required equation of the circle is

⇒ (x – 4)^{2} + (y – 0)^{2} = 4^{2} or (x + 4)^{2} + (y – 0)^{2} = 4^{2}

⇒ x^{2} – 8x + 16 + y^{2} = 16 or x^{2} + 8x + 16 + y^{2} = 16

⇒ x^{2} + y^{2} – 8x = 0 or x^{2} + y^{2} + 8x = 0

(iv) Centre of the circle is C (3, 1).

Let the circle touch the line 8x – 15y + 25 = 0 at point M.

CM = radius (r)

CM = Length of perpendicular from centre C(3, 1) on the line 8x – 15y + 25 = 0

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

(x – h)^{2} + (y – k)^{2} = r^{2}

Here, h = 3, k = 1 and r = 2

The required equation of the circle is

⇒ (x – 3)^{2} + (y – 1)^{2} = 2^{2}

⇒ x^{2} – 6x + 9 + y^{2} – 2y + 1 = 4

⇒ x^{2} + y^{2} – 6x – 2y + 10 – 4 = 0

⇒ x^{2} + y^{2} – 6x – 2y + 6 = 0

Question 4.

Find the equation of the circle, if the equations of two diameters are 2x + y = 6 and 3x + 2y = 4 and radius is 9.

Solution:

Given equations of diameters are 2x + y = 6 and 3x + 2y = 4.

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

Since point of intersection of diameters is the centre of the circle,

x = h, y = k

Equations of diameters become

2h + k = 6 …..(i)

and 3h + 2k = 4 ……..(ii)

By (ii) – 2 × (i), we get

-h = -8

⇒ h = 8

Substituting h = 8 in (i), we get

2(8) + k = 6

⇒ k = 6 – 16

⇒ k = -10

Centre of the circle is C (8, -10) and radius, r = 9

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

(x – h)^{2} + (y – k)^{2} = r^{2}

Here, h = 8, k = -10

The required equation of the circle is

⇒ (x – 8)^{2} + (y + 10)^{2} = 9^{2}

⇒ x^{2} – 16x + 64 + y^{2} + 20y + 100 = 81

⇒ x^{2} + y^{2} – 16x + 20y + 100 + 64 – 81 = 0

⇒ x^{2} + y^{2} – 16x + 20y + 83 = 0

Question 5.

If y = 2x is a chord of the circle x^{2} + y^{2} – 10x = 0, find the equation of the circle with this chord as diameter.

Solution:

y = 2x is the chord of the given circle.

It satisfies the equation of a given circle.

Substituting y = 2x in x^{2} + y^{2} – 10x = 0, we get

⇒ x^{2} + (2x)^{2} – 10x = 0

⇒ x^{2} + 4x^{2} – 10x = 0

⇒ 5x^{2} – 10x = 0

⇒ 5x(x – 2) = 0

⇒ x = 0 or x = 2

When x = 0, y = 2x = 2(0) = 0

∴ A = (0, 0)

When x = 2, y = 2x = 2 (2) = 4

∴ B = (2, 4)

End points of chord AB are A(0, 0) and B(2, 4).

Chord AB is the diameter of the required circle.

The equation of a circle having (x_{1}, y_{1}) and (x_{2}, y_{2}) as end points of diameter is given by

(x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

Here, x_{1} = 0, y_{1} = 0, x_{2} = 2, y_{2} = 4

The required equation of the circle is

⇒ (x – 0) (x – 2) + (y – 0) (y – 4 ) = 0

⇒ x^{2} – 2x + y^{2} – 4y = 0

⇒ x^{2} + y^{2} – 2x – 4y = 0

Question 6.

Find the equation of a circle with a radius of 4 units and touch both the co-ordinate axes having centre in the third quadrant. Solution:

The radius of the circle = 4 units

Since the circle touches both the co-ordinate axes and its centre is in the third quadrant,

the centre of the circle is C(-4, -4).

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

Here, h = -4, k = -4, r = 4

the required equation of the circle is

⇒ [x – (-4)]^{2} + [y – (-4)]^{2} = 4^{2}

⇒ (x + 4)^{2} + (y + 4)^{2} = 16

⇒ x^{2} + 8x + 16 + y^{2} + 8y + 16 – 16 = 0

⇒ x^{2} + y^{2} + 8x + 8y + 16 = 0

Question 7.

Find the equation of the circle passing through the origin and having intercepts 4 and -5 on the co-ordinate axes.

Solution:

Let the circle intersect X-axis at point A and intersect Y-axis at point B.

the co-ordinates of point A are (4, 0) and the co-ordinates of point B are (0, -5).

Since ∠AOB is a right angle,

AB represents the diameter of the circle.

The equation of a circle having (x_{1}, y_{1}) and (x_{2}, y_{2}) as end points of diameter is given by

(x – x_{1}) (x – x_{2}) + (y – y_{1}) (y – y_{2}) = 0

Here, x_{1} = 4, y_{1} = 0, x_{2} = 0, y_{2} = -5

The required equation of the circle is

⇒ (x – 4) (x – 0) + (y – 0) [y – (-5)] = 0

⇒ x(x – 4) + y(y + 5) = 0

⇒ x^{2} – 4x + y^{2} + 5y = 0

⇒ x^{2} + y^{2} – 4x + 5y = 0

Question 8.

Find the equation of a circle passing through the points (1, -4), (5, 2) and having its centre on line x – 2y + 9 = 0.

Solution:

Let C(h, k) be the centre of the required circle which lies on the line x – 2y + 9 = 0.

Equation of line becomes

h – 2k + 9 = 0 …..(i)

Also, the required circle passes through points A(1, -4) and B(5, 2).

CA = CB = radius

CA = CB

By distance formula,

\(\sqrt{(\mathrm{h}-1)^{2}+[\mathrm{k}-(-4)]^{2}}=\sqrt{(\mathrm{h}-5)^{2}+(\mathrm{k}-2)^{2}}\)

Squaring both the sides, we get

⇒ (h – 1)^{2} + (k + 4)^{2} = (h – 5)^{2} + (k – 2)^{2}

⇒ h^{2} – 2h + 1 + k^{2} + 8k + 16 = h^{2} – 10h + 25 + k^{2} – 4k + 4

⇒ -2h + 8k + 17 = -10h – 4k + 29

⇒ 8h + 12k – 12 = 0

⇒ 2h + 3k – 3 = 0 ……(ii)

By (ii) – (i) × 2, we get

7k = 21

⇒ k = 3

Substituting k = 3 in (i), we get

h – 2(3) + 9 = 0

⇒ h – 6 + 9 = 0

⇒ h = -3

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

radius (r) = CA

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

Here, h = -3, k = 3, r = √65

The required equation of the circle is

⇒ [x – (-3)]^{2} + (y – 3)^{2} = (√65)^{2}

⇒ (x + 3)^{2} + (y – 3)^{2} = 65

⇒ x^{2} + 6x + 9 + y^{2} – 6y + 9 – 65 = 0

⇒ x^{2} + y^{2} + 6x – 6y – 47 = 0