Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 6 Circle Ex 6.2 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 6 Circle Ex 6.2

Question 1.

Find the centre and radius of each of the following circles:

(i) x^{2} + y^{2} – 2x + 4y – 4 = 0

(ii) x^{2} + y^{2} – 6x – 8y – 24 = 0

(iii) 4x^{2} + 4y^{2} – 24x – 8y – 24 = 0

Solution:

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

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

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

⇒ g = -1, f = 2 and c = -4

Centre of the circle = (-g, -f) = (1, -2)

and radius of the circle

(ii) Given equation of the circle is x^{2} + y^{2} – 6x – 8y – 24 = 0

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

2g = -6, 2f = -8 and c = -24

⇒ g = -3, f = -4 and c = -24

Centre of the circle = (-g, -f) = (3, 4)

and radius of the circle

(iii) Given equation of the circle is 4x^{2} + 4y^{2} – 24x – 8y – 24 = 0

Dividing throughout by 4, we get x^{2} + y^{2} – 6x – 2y – 6 = 0

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

2g = -6, 2f = -2 and c = -6

⇒ g = -3, f = -1 and c = -6

Centre of the circle = (-g, -f) = (3, 1)

and radius of the circle

Question 2.

Show that the equation 3x^{2} + 3y^{2} + 12x + 18y – 11 = 0 represents a circle.

Solution:

Given equation is 3x^{2} + 3y^{2} + 12x + 18y – 11 = 0

Dividing throughout by 3, we get

x^{2} + y^{2} + 4x + 6y – \(\frac{11}{3}\) = 0

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

2g = 4, 2f = 6, c = \(\frac{-11}{3}\)

⇒ g = 2, f = 3, c = \(\frac{-11}{3}\)

Now, g^{2} + f^{2} – c = (2)^{2} + (3)^{2} – (\(\frac{-11}{3}\))

= 4 + 9 + \(\frac{11}{3}\)

= \(\frac{50}{3}\) > 0

∴ The given equation represents a circle.

Question 3.

Find the equation of the circle passing through the points (5, 7), (6, 6), and (2, -2).

Solution:

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

Since the required circle passes through points A(5, 7), B(6, 6), and D(2, -2),

CA = CB = CD = radius

Consider, CA = CD

By distance formula,

\(\sqrt{(\mathrm{h}-5)^{2}+(\mathrm{k}-7)^{2}}=\sqrt{(\mathrm{h}-2)^{2}+[\mathrm{k}-(-2)]^{2}}\)

Squaring both the sides, we get

⇒ (h – 5)^{2} + (k – 7)^{2} = (h – 2)^{2} + (k + 2)^{2}

⇒ h^{2} – 10h + 25 + k^{2} – 14k + 49 = h^{2} – 4h + 4 + k^{2} + 4k + 4

⇒ -10h – 14k + 74 = -4h + 4k + 8

⇒ 6h + 18k – 66 = 0

⇒ h + 3k – 11 = 0 …..(i)

Consider, CB = CD

By distance formula,

\(\sqrt{(h-6)^{2}+(k-6)^{2}}=\sqrt{(h-2)^{2}+[k-(-2)]^{2}}\)

Squaring both the sides, we get

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

⇒ h^{2} – 12h + 36 + k^{2} – 12k + 36 = h^{2} – 4h + 4 + k^{2} + 4k + 4

⇒ -12h – 12k + 72 = -4h + 4k + 8

⇒ 8h + 16k – 64 = 0

⇒ h + 2k – 8 = 0 ……(ii)

By (i) – (ii), we get k = 3

Substituting k = 3 in (i), we get

h + 3(3) – 11 = 0

⇒ h + 9 – 11 = 0

⇒ h = 2

Centre of the circle is C(2, 3).

radius (r) = CD

= \(\sqrt{(2-2)^{2}+(3+2)^{2}}\)

= \(\sqrt{0+5^{2}}\)

= √25

= 5

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} = 5^{2}

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

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

⇒ x^{2} + y^{2} – 4x – 6y – 12 = 0

Question 4.

Show that the points (3, -2), (1, 0), (-1, -2) and (1, -4) are concyclic.

Solution:

Let the equation of the circle passing through the points (3, -2), (1, 0) and (-1, -2) be

x^{2} + y^{2} + 2gx + 2fy + c = 0 …..(i)

For point (3, -2),

Substituting x = 3 and y = -2 in (i), we get

9 + 4 + 6g – 4f + c = 0

⇒ 6g – 4f + c = -13 ….(ii)

For point (1, 0),

Substituting x = 1 andy = 0 in (i), we get

1 + 0 + 2g + 0 + c = 0

⇒ 2g + c = -1 ……(iii)

For point (-1, -2),

Substituting x = -1 and y = -2, we get

1 + 4 – 2g – 4f + c = 0

⇒ 2g + 4f – c = 5 …….(iv)

Adding (ii) and (iv), we get

8g = -8

⇒ g = -1

Substituting g = -1 in (iii), we get

-2 + c = -1

⇒ c = 1

Substituting g = -1 and c = 1 in (iv), we get

-2 + 4f – 1 = 5

⇒ 4f = 8

⇒ f = 2

Substituting g = -1, f = 2 and c = 1 in (i), we get

x^{2} + y^{2} – 2x + 4y + 1 = 0 ……….(v)

If (1, -4) satisfies equation (v), the four points are concyclic.

Substituting x = 1, y = -4 in L.H.S of (v), we get

L.H.S. = (1)^{2} + (-4)^{2} – 2(1) + 4(-4) + 1

= 1 + 16 – 2 – 16 + 1

= 0

= R.H.S.

Point (1, -4) satisfies equation (v).

∴ The given points are concyclic.