Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.1 Textbook Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 5 Two Dimensional Analytical Geometry – II Ex 5.1

Question 1.

Obtain the equation of the circles with a radius of 5 cm and touching the x-axis at the origin in a general form.

Solution:

Given radius = 5 cm and the circle is touching x axis

So centre will be (0, ± 5) and radius = 5

The equation of the circle with centre (0, ± 5) and radius 5 units is

(x – 0)^{2} + (y ± 5)^{2} = 5^{2}

(i.e) x^{2} + y^{2} ± 10 y + 25 – 25 = 0

(i.e) x^{2} + y^{2} ± 10y = 0

Question 2.

Find the equation of the circle with centre (2, – 1) and passing through the point (3, 6) in standard form.

Solution:

Centre = (2, -1) = (h, k)

Passing through the point (3, 6)

Equation of the circle (x – h)² + (y – k)² = r² ………. (1)

(3 – 2)² + (6 + 1)² = r²

1² + 7² = r²

1 + 49 = r²

r² = 50

(1) ⇒ (x – 2)² + (y + 1)² = 50

Question 3.

Find the equation of circles that touch both the axes and pass-through (-4, -2) in a general form.

Solution:

Since the circle touches both the axes, its centre will be (r, r) and the radius will be r.

Here centre = C = (r, r) and point on the circle is A = (-4, -2)

CA = r ⇒ CA^{2} = r^{2}

(i.e) (r + 4)^{2} + (r + 2)^{2} = r^{2}

⇒ r^{2} + 8r +16 + r^{2} + 4r + 4 – r^{2} = 0

(i.e) r^{2} + 12r + 20 = 0

(r + 2) (r + 10) = 0

⇒ r = -2 or -10

When r = -2, the equation of the circle will be (x + 2)^{2} + (y + 2)^{2} = 2^{2}

(i.e) x^{2} + y^{2} + 4x + 4y + 4 = 0

When r = -10, the equation of the circle will be (x + 10)^{2} + (y + 10)^{2} = 10^{2}

(i.e) x^{2} + y^{2} + 20x + 20y + 100 = 0

Question 4.

Find the equation of the circles with centre (2, 3) and passing through the intersection of the lines 3x – 2y – 1 =0 and 4x + y – 27 = 0.

Solution:

centre (2, 3) = (h, k)

Point of intersection

Solve 3x – 2y – 1 = 0 ………. (1)

4x + y – 27 = 0 ……… (2)

(1) ⇒ 3x – 2y = 1

(2) × 2 ⇒ 8x + 2y = 54

11x = 55

x = 5

put in (1)

15 – 2y – 1 = 0

14 = 2y

y = 7

Passing-through point is (5, 7)

Equation of circle be (x – h)² + (y – k)² = r² ……….(3)

(5 – 2)² + (7 – 3)² = r²

3² + 4² = r²

r² = 25

∴ (3) ⇒ (x – 2)² + (y – 3)² = 25

x² – 4x + 4 + y² – 6y + 9 – 25 = 0

x² + y² – 4x – 6y – 12 = 0

Question 5.

Obtain the equation of the circle for which (3, 4) and (2, -7) are the ends of a diameter.

Solution:

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

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

Here the end points of a diameter are (3, 4) and (2, -7)

So equation of the circle is (x – 3 )(x – 2 ) + (y – 4) (y + 7.) = 0

x^{2} + y^{2} – 5x + 37 – 22 = 0

Question 6.

Find the equation of the circle through the points (1, 0), (-1, 0) and (0, 1).

Solution:

Let the general equation of the circle be

x² + y² + 2gx + 2fy + c = 0

It passes through the points (1, 0), (-1, 0) and (0,1)

1 + 0 + 2g + c = 0

2g + c = -1 ………(1)

1 + 0 – 2g + c = 0

-2g + c = -1 …………(2)

0 + 1 + 0 + 2f + c = 0

2f + c = -1

(1) + (2) ⇒ 2c = -2

c = -1

substitute in eqn (1)

2g – 1 = -1

2g = 0

g = 0

substitute in eqn (3)

2f – 1 = -1

2f = -1 + 1

2f = 0

f = 0

Therefore, the required equation of the circle

x² + y² – 1 = 0

Question 7.

A circle of area 9π square units has two of its diameters along the lines x + y = 5 and: x – y = 1. Find the equation of the circle.

Solution:

Area of the circle = 9π

(i.e) πr^{2} = 9π

⇒ r^{2} = 9 ⇒ r = 3

(i.e) radius of the circle = r = 3

The two diameters are x + y = 5 and x – y = 1

The point of intersection of the diameter is the centre of the circle = C

To find C: Solving x + y = 5 ……… (1)

x – y = 1 ………. (2)

(1) + (2) ⇒ 2x = 6 ⇒ x = 3

Substituting x = 3 in (1) we get

3 + y = 5 ⇒ y = 5 – 3 = 2

∴ Centre = (3, 2) and radius = 3

So equation of the circle is (x – 3)^{2} + (y – 2)^{2} = 3^{2}

(i.e) x^{2} + y^{2} – 6x – 4y + 4 = 0

Question 8.

If y = 2√2 x + c is a tangent to the circle x² + y² = 16, find the value of c.

Solution:

The condition of the line y = mx + c to be a tangent to the circle x² + y² = a² is

c² = a²( 1 + m²)

a² = 16; m = 2√2 ⇒ m² = 4 × 2 = 8

c² = 16(1+8)

c² = 16(9)

c = ±4 × 3 = ±12

∴ c = ± 12.

Question 9.

Find the equation of the tangent and normal to the circle x² + y² – 6x + 6y – 8 = 0 at (2, 2).

Solution:

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

xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c = 9

So the equation of the tangent to the circle

x^{2} + y^{2} – 6x + 6y – 8 = 0 at (x_{1}, y_{1}) is

xx_{1} + yy_{1} – \(\frac{6\left(x+x_{1}\right)}{2}+\frac{6\left(y+y_{1}\right)}{2}\) – 8 = 0

(i.e) xx_{1} + yy_{1} – 3(x + x_{1}) + 3(y + y_{1}) – 8 = 0

Here (x_{1}, y_{1}) = (2, 2)

So equation of the tangent is

x(2) + y(2) – 3(x + 2) + 3(y + 2) – 8 = 0

(.i.e) 2x + 2y – 3x – 6 + 3y + 6 – 8 = 0

(i.e) -x + 5y – 8 = 0 or x – 5y + 8=0

Normal is a line ⊥r to the tangent

So equation of normal circle be of the form 5x + y + k = 0

The normal is drawn at (2, 2)

⇒ 10 + 2 + k = 0 ⇒ k = -12

So equation of normal is 5x + y – 12 = 0

Question 10.

Determine whether the points (- 2, 1), (0, 0) and s (- 4, – 3) lie outside, on or inside the circle x² + y² – 5x + 2y – 5 = 0.

Solution:

x² + y² – 5x + 2y – 5 = 0

(i) At (-2, 1) ⇒ (-2)² + 1² – 5(-2) + 2(1) – 5

= 4 + 1 + 10 + 2 – 5 = 12 > 0

∴ (-2, 1) lies outside the circle.

(ii) At(0, 0) ⇒ 0 + 0 – 0 + 0 – 5 = -5 < 0

(0, 0) lies inside the circle.

(iii) At (-4, -3) ⇒ (-4)² + (- 3)² – 5(-4) + 2(-3) – 5 = 16 + 9 + 20 – 6 – 5 = 34 > 0

(-4, -3) lies outside the circle.

Question 11.

Find the centre and radius of the following circles.

(i) x² + (y + 2)² = 0

(ii) x² + y² + 6x – 4y + 4 = 0

(iii) x² + y² – x + 2y – 3 = 0

(iv) 2x² + 2y² – 6x + 4y + 2 = 0

Solution:

(i) x^{2} + (y + 2)^{2} = 0

(i.e) x^{2} + y^{2} + 4y + 4 = 0

Comparing this equation with the general form x^{2} + y^{2} + 2gx + 2fy + c = 0

we get 2g = 0 ⇒ g = 0

2f = 4 ⇒ f= 2 and c = 4

Now centre = (-g, -f) = (0, -2)

Radius = r = \(\sqrt{g^{2}+f^{2}-c}=\sqrt{0+4-4}\)

∴ Centre = (0, -2) and radius = 0

(ii) x^{2} + y^{2} + 6x – 4y + 4 = 0

Comparing with the general form we get

2g = 6, 2f = -4

⇒ g = 3, /= -2 and c = 4

Centre = (-g, -f) = (-3, 2)

Radius = \(\sqrt{g^{2}+f^{2}-c}=\sqrt{9+4-4}\)= 3

∴ Centre = (-3, 2) and radius = 3

(iii) x² + y² – x + 2y – 3 = 0

2g = -1; 2f = 2; c = -3

g = \(\frac{-1}{2}\) f = 1

Centre (-g, -f) = (\(\frac{1}{2}\), -1)

Radius = \(\sqrt{g^2+f^2-c}\) = \(\sqrt{\frac{1}{4}+1+3}\)

\(\sqrt{\frac{1+4+12}{4}}\)

\(\sqrt{\frac{17}{4}}\) = \(\frac{\sqrt{17}}{2}\)

(iv) 2x^{2} + 2y^{2} – 6x + 4y + 2 = 0

(÷ by 2) ⇒ x^{2} + y^{2} – 3x + 2y + 1 =0

Comparing this equation with the general form of the circle we get

2g = -3, 2f= 2

g = \(-\frac{3}{2}\), g= 1 and c = 1

So centre = (-g, -f) = (\(\frac{3}{2}\), -1)

and radius = \(\sqrt{g^{2}+f^{2}-c}=\sqrt{\frac{9}{4}+1-1}=\frac{3}{2}\)

∴ Centre = (\(\frac{3}{2}\), -1) and radius = \(\frac{3}{2}\)

Question 12.

If the equation 3x² + (3 – p) xy + qy² – 2px = 8pq represents a circle, find p and q. Also determine the centre and radius of the circle.

Solution:

3x² + (3 – p) xy + qy² – 2px = 8pq represent a circle means,

Co-efficient of x² = co-efficient of y²

3 = q ⇒ q = 3

Co-efficient of xy = 0

3 – p = 0 ⇒ p = 3

3x² + 3y² – 6x = 8 (3)(3)

3x² + 3y² – 6x – 72 = 0

(÷3) x² + y² – 2x – 24 = 0

2g = -2; 2f = 0; c = -24

g = -1 f = 0

Centre (-g, -f) = (1, 0)

Radius = \(\sqrt{g^2+f^2-c}\) = \(\sqrt{1+0+24}\)

= \(\sqrt{25}\) = 5