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

**12th Maths Exercise 5.1 Samacheer Kalvi Question 1.**

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

**12th Maths Exercise 5.1 Question 2.**

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

Solution:

Centre = C = (2, -1); Passing through = A = (3, 6)

So radius = CA = \(\sqrt{(2-3)^{2}+(-1-6)^{2}}=\sqrt{1+49}=5 \sqrt{50}\)

Now centre = (2, -1) and radius = \(\sqrt{50}\)

So equation of the circle is

(i.e) (x – 2)^{2} + (y + 1)^{2} = \(\sqrt{50}^{2}\) ⇒ (x – 2)^{2} + (y + 1)^{2} = 50

**12th Maths Exercise 5.2 Samacheer Kalvi Question 3.**

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

Solution:

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

**12th Maths Chapter 5 Exercise 5.1 Question 4.**

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

Solution:

To find the point of intersection of the two lines we have to solve the two equations.

Now solving them: 3x — 2y = 1 ……….. (1)

4x + y = 27 ……….(2)

(2) × 2 ⇒ 8x + 2y = 54 ……….. (3)

(1) ⇒ 3x – 2y = 1 ………. (1)

(3) + (1) ⇒ 11x = 55 ⇒ x = \(\frac{55}{11}\) = 5

Substituting x = 5 in (2) we get

20 + y = 27 ⇒ y = 27 – 20 = 7

∴ The point = A = (5, 7)

Given centre = C = (2, 3)

∴ radius = \(\sqrt{(5-2)^{2}+(7-3)^{2}}=\sqrt{9+16}=5\) = 5

Now centre = (2, 3) and radius = 5

So equation of the circle is

(x – 2)^{2} + (y – 3)^{2} = 5^{2}

(i.e) x^{2} + y^{2} – 4x – 6y – 12 =0

Class 12 Maths Chapter 5 Exercise 5.1 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

**Two Dimensional Analytical Geometry 2 Question 6.**

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

Solution:

Let the required circle be

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

The circle passes through (1, 0), (-1, 0) and (0, 1)

(1, 0) ⇒ 1 + 0 + 2g(1) + 2f(0) + c = 0

2g + c = -1 ……………. (1)

(-1, 0) ⇒ 1 + 0 + 2g (-1) + 2f(0) + c = 0

-2g + c = -1 ……….. (2)

(0, 1) ⇒ 0 + 1 + 2g (0) + 2f(1) + c = 0

2f+ c = -1 ……….. (3)

Now solving (1), (2) and (3) .

2g + c = -1 ………… (1)

-2g + c = -1 ………….. (2)

(1) + (2) ⇒ 2c = -2 ⇒ c = -1

Substituting c = -1 in (1) we get

2g – 1 = -1

2g = -1 + 1 = 0 ⇒ g = 0

Substituting c = -1 in (3) we get

2f – 1 =-1 ⇒ 2f = -1 + 1 = 0 ⇒ f = 0

So we get g = 0, f= 0 and c = -1

So the required circle will be

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

(i.e) x^{2} + y^{2} – 1 = 0 ⇒ x^{2} + y^{2} = 1

**Exercise 5.1 Class 12 Question 7.**

A circle of area 9n 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

**Ex 5.1 Class 12 Question 8.**

If y = \(2 \sqrt{2} x\) + c is a tangent to the circle x^{2} + y^{2} =16 , find the value of c .

Solution:

The condition for the line y = mx + c to be a tangent to the circle x^{2} + y^{2} = a^{2} is

c^{2} = a^{2}(1 + m^{2})

Here x^{2} + y^{2} = 16 ⇒ a^{2} = 16

y = \(2 \sqrt{2} x\) + c ⇒ m = \(2 \sqrt{2}\) and c = c

The condition is c^{2} = a^{2} (1 + m^{2})

(i.e) c^{2} = 16(1 + 8) = 144

⇒ c = ± 12

**Chapter 5 Maths Class 12 Question 9.**

Find the equation of the tangent and normal to the circle x^{2} + y^{2} – 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

**Chapter 5 Class 12 Maths Question 10.**

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

Solution:

To find the position of a point with regard to a given circle, substitute the point in the equation of the circle if we get a positive value, the point lies outside the circle.

If we get a -ve value the point lies inside the circle and if we get O then the point lies on the circumference of the circle.

The given circle is x^{2} + y^{2} – 5x + 2y – 5 = 0 ……….. (1)

Substituting the point (-2, 1) in (1) we get

4 + 1 – 5(-2) + 2(1) – 5 = 5 + 10 + 2 – 5 = 12

⇒ (- 2, 1) lies outside the circle

Substituting the point (0, 0) in (1) we get

-5 < 0 ⇒ (0, 0) lies inside the circle Substituting the point (-4, -3) in (1) we get 16 + 9 + 20 – 6 – 5 = 34 >0

⇒ (- 4, -3) lies outside the circle

**12th Maths 5th Chapter Samacheer Kalvi Question 11.**

Find centre and radius of the following circles.

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

(ii) x^{2} + y^{2} + 6x – 4y + 4 = 0

(iii) x^{2} + y^{2} – x + 2y – 3 = 0

(iv) 2x^{2} + 2y^{2} – 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) x^{2} + y^{2} – x + 2y – 3 = 0

Comparing with the general form of the circle we get

(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 = \(\frac{3}{2}\)

**Two Dimensional Analytical Geometry Pdf Question 12.**

If the equation 3x^{2} + (3 -p) xy + qy^{2} – 2px = 8 pq represents a circle, find p and q. Also determine the centre and radius of the circle.

Solution:

For a circle co-eff of x^{2} = co-efif of y^{2}

⇒ 3 = q

co-eff of xy = 0

⇒ 3 – p = 0 ⇒ p = 3

So p = q = 3

So the equation of the circle becomes 3x^{2} + 3y^{2} – 6x – 72 = 0

(÷ by 3) ⇒ x^{2} + y^{2} – 2x – 24 = 0

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

2g = -2, 2f = 0

g = -1, f = 0 and c = -24

So centre = (-g, -f) = (1, 0) and radius = \(\sqrt{g^{2}+f^{2}-c}\) = 5

∴ Centre =(1,0) and radius = 5

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

**Ex5 1 Class 12 Question 1.**

Find the equation of the circle whose centre is (2, -3) and passing through the intersection of the line 3x – 2y = 1 and 4x + y = 27.

Solution:

Solving 3x – 2y = 1 and 4x+y = 27

Simultaneously, we get x = 5 and y = 7

∴ The point of intersection of the lines is (5, 7)

Now we have to find the equation of a circle whose centre is (2, -3) and which passes through (5,7)

∴ Required equation of the circle is

(x – 2)^{2} + (y + 3)^{2} = \((\sqrt{109})^{2}\)

⇒ x^{2} + y^{2} – 4x + 6y – 96 = 0

**Class 12 Maths Chapter 5 Question 2.**

Find the equation of a circle of radius 5 whose centre lies on x-axis and which passes through the point (2, 3).

Solution:

Let the centre of the circle be (h, k). Since the centre lies on x-axis, we have k = 0

Therefore, centre = (h, 0)

Also the circle has radius 5 units and passes through the point (2, 3).

Hence the centre is (6, 0) or (-2, 0)

∴ The equation of the circle is

(x – 6)^{2} + (y – 0)^{2} = 5^{2} ⇒ x^{2} + y^{2} – 12x+ 11 = 0 (Or) (x + 2)^{2} + (y – 0)^{2} = 5^{2} ⇒ x^{2} + y^{2} + 4x – 21 =0

**Two Dimensional Analytical Geometry Question 3.**

Find the centre and radius of the following circles:

x^{2} + y^{2} – 2x + 4y – 4 = 0 and 2x^{2} + 2y^{2} + 16x – 28y + 32 = 0. Also find the ratio of their diameters.

Solution:

Comparing the equation x^{2} + y^{2} – 2x + 4y – 4 = 0 with x^{2} + y^{2} + 2gx + 2 fy + c = 0, we get

2g = -2 ⇒ g = -1, 2f = 4 ⇒ f = 2, c = -4

Comparing with x^{2} + y^{2} + 2gx + 2fy + c = 0

2g = 8 ⇒ g = 4, 2f = -14 ⇒ f = -7, c = 16

Diameter = 2r_{2} = 2 × 7 = 14 units

Ratio of their diameters = 6 : 14 = 3 : 7.

**Class 12 Maths Ex5.1 Question 4.**

Find the equation of the circle whose radius is 4 and which is concentric with the circle x^{2} + y^{2} + 2x – 6y = 0

Solution:

x^{2} + y^{2} + 2x – 6y = 0 …(1)

Here 2g = 2 ⇒ g = 1, 2f = -6 ⇒ f = -3

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

Since the required circle is concentric with (1), its centre is also (-1, 3).

∴ The equation of the circle whose centre is (-1, 3) and radius 4 is

(x + 1)^{2} + (y – 3)^{2} = 4^{2}

⇒ x^{2} + y^{2} + 2x – 6y – 6 = 0

**Exercise 5.1 Class 12 Solutions Question 5.**

Show that the four points (1, 0), (2, -7), (8,1) and (9, 6) are concyclic.

Solution:

Let the equation of the circle be Since it passes through (1,0) ⇒ x^{2} + y^{2} + 2gx + 2fy + c = 0 …(1)

**Exercise 5.1 Class 12 Maths Question 6.**

Find the equation of a circle which passes through the points (1, -2) and (4, -3) and whose centre lies on the line 3x + 4y = 0

Solution:

Let the equation of the circle be

x^{2} + y^{2} + 2gx + 2fy + c = 0 …(1)

Since it passes through (1, -2) and (4, -3)

∴ 5 + 2g – 4f + c = 0 …(2)

and 25 + 8g – 6f + c = 0 …(3)

Also (-g, -f) centre of circle (1) lies on 3x + 4y = 7

-3g – 4f = 7 …(4)

Subtracting (2) from (3), we get

20 + 6g – 2f = 0 …(5)

Solving (4) and (5), we get

Substituting these values of g and f in (2)

∴ From (1), we get,

Which is the required equation on the circle.

**Math Solution Class 12 Chapter 5 Question 7.**

Find the equation (s) of the circle passing through the points (1, 1) and (2, 2) and whose radius is 1.

Solution:

Let the equation of the circle be

x^{2} + y^{2} + 2gx + 2fy + c = 0 …. (1)

Since it passes through (1,1)

∴ 1 + 1 + 2g + 2f + c = 0

⇒ 2g + 2f + c + 2 = 0 ….(2)

Again it passes through (2, 2)

∴ 4 + 4+ 4g + 4/+ c = 0

⇒ 4g + 4f + c + 8 = 0 …(3)

Adding (2) and (4)

g^{2} + 2g + f^{2} + 2f + 2 = 1

g^{2} + 2g + (-g – 3)^{2} + 2(-g – 3) + 1 = 0

⇒ 2g^{2} + 6g + 4 = 0 ⇒ g^{2} + 3g + 2 = 0

⇒ (g + 1)(g + 2) = 0 ⇒ g = -1, -2

when g = -1, from (5), f = – (-1) – 3 = 1 – 3 = -2

when g = -2, from (5), f = – (-2) – 3 = 2 – 3 = -1

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

2(-1) + 2(-2) + c + 2 =0

⇒ – 6 + c + 2 = 0 ⇒ c = 4

Now putting g = -1, f = -2 and c = 4 in (1), we get

x^{2} + y^{2} – 2x – 4y + 4 = 0

Again putting g = -2, f = -1 in (2), we get

2(-2) + 2(-1) + c + 2 =0

⇒ -4 – 2 + c + 2 =0 ⇒ c = 4

Putting g = -2, f = -1 and c = 4 (in) (1), we get

x^{2} + y^{2} – 4x – 2y + 4 = 0

Hence the required equation (s) of the circle are

x^{2} + y^{2} – 2x – 4y + 4 =0 and x^{2} + y^{2} – 4x – 2y + 4 = 0.