Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 6 Circle Ex 6.3 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 6 Circle Ex 6.3

Question 1.

Write the parametric equations of the circles:

(i) x^{2} + y^{2} = 9

(ii) x^{2} + y^{2} + 2x – 4y – 4 = 0

(iii) (x – 3)^{2} + (y + 4)^{2} = 25

Solution:

(i) Given equation of the circle is

x^{2} + y^{2} = 9

⇒ x^{2} + y^{2} = 3^{2}

Comparing this equation with x^{2} + y^{2} = r^{2}, we get r = 3

The parametric equations of the circle in terms of θ are

x = r cos θ and y = r sin θ

⇒ x = 3 cos θ and y = 3 sin θ

(ii) Given equation of the circle is

x^{2} + y^{2} + 2x – 4y – 4 = 0

⇒ x^{2} + 2x + y^{2} – 4y – 4 = 0

⇒ x^{2} + 2x + 1 – 1 + y^{2} – 4y + 4 – 4 – 4 = 0

⇒ (x^{2} + 2x + 1 ) + (y^{2} – 4y + 4) – 9 = 0

⇒ (x + 1)^{2} + (y – 2)^{2} = 9

⇒ (x + 1)^{2} + (y – 2)^{2} = 3^{2}

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

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

The parametric equations of the circle in terms of θ are

x = h + r cos θ and y = k + r sin θ

⇒ x = -1 + 3 cos θ and y = 2 + 3 sin θ

(iii) Given equation of the circle is

(x – 3)^{2} + (y + 4)^{2} = 25

⇒ (x – 3)^{2} + (y + 4)^{2} = 5^{2}

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

h = 3, k = -4 and r = 5

The parametric equations of the circle in terms of θ are

x = h + r cos θ and y = k + r sin θ

⇒ x = 3 + 5 cos θ and y = -4 + 5 sin θ

Question 2.

Find the parametric representation of the circle 3x^{2} + 3y^{2} – 4x + 6y – 4 = 0.

Solution:

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

Dividing throughout by 3, we get

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

h = \(\frac{2}{3}\), k = -1 and r = \(\frac{5}{3}\)

The parametric representation of the circle in terms of θ are

x = h + r cos θ and y = k + r sin θ

⇒ x = \(\frac{2}{3}\) + \(\frac{5}{3}\) cos θ and y = -1 + \(\frac{5}{3}\) sin θ

Question 3.

Find the equation of a tangent to the circle x^{2} + y^{2} – 3x + 2y = 0 at the origin.

Solution:

Given equation of the circle is x^{2} + y^{2} – 3x + 2y = 0

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0, we get

2g = -3, 2f = 2, c = 0

⇒ g = \(-\frac{3}{2}\), f = 1, c = 0

The equation of a tangent to the circle

x^{2} + y^{2} + 2gx + 2fy + c = 0 at (x_{1}, y_{1}) is xx_{1} +yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c = 0

The equation of the tangent at (0, 0) is

x(0) + y(0) + (\(-\frac{3}{2}\)) (x + 0) + 1(y + 0) + 0 = 0

⇒ \(-\frac{3}{2}\)x + y = 0

⇒ 3x – 2y = 0

Question 4.

Show that the line 7x – 3y – 1 = 0 touches the circle x^{2} + y^{2} + 5x – 7y + 4 = 0 at point (1, 2).

Solution:

Given equation of the circle is x^{2} + y^{2} + 5x – 7y + 4 = 0

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0, we get

2g = 5, 2f = -7, c = 4

⇒ g = \(\frac{5}{2}\), f = \(\frac{-7}{2}\), c = 4

The equation of a tangent to the circle x^{2} + y^{2} + 2gx + 2fy + c = 0 at (x_{1}, y_{1}) is

xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c = 0

The equation of the tangent at (1, 2) is

7x – 3y – 1 = 0, which is same as the given line.

The line 7x – 3y – 1 = 0 touches the given circle at (1, 2).

Question 5.

Find the equation of tangent to the circle x^{2} + y^{2} – 4x + 3y + 2 = 0 at the point (4, -2).

Solution:

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

Comparing this equation with x^{2} + y^{2} + 2gx + 2fy + c = 0, we get

2g = -4, 2f = 3, c = 2

g = -2, f = \(\frac{3}{2}\), c = 2

The equation of a tangent to the circle x^{2} + y^{2} + 2gx + 2fy + c = 0 at (x_{1}, y_{1}) is

xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c = 0

The equation of the tangent at (4, -2) is

x(4) + y(-2) – 2(x + 4) + \(\frac{3}{2}\)(y – 2) + 2 = 0

⇒ 4x – 2y – 2x – 8 + \(\frac{3}{2}\) y – 3 + 2 = 0

⇒ 2x – \(\frac{1}{2}\)y – 9 = 0

⇒ 4x – y – 18 = 0