Find the equation of the circle with:

(i) Centre (-2, 3) and radius 4.

(ii) Centre (a, b) and radius .

(iii) Centre (0, – 1) and radius 1.

(iv) Centre (a cos α, a sin α) and radius a.

(v) Centre (a, a) and radius √2 a.

**Answer
1** :

(i) Centre (-2, 3) andradius 4.

Given:

The radius is 4 andthe centre (-2, 3)

By using the formula,

The equation of thecircle with centre (p, q) and radius ‘r’ is (x – p)^{2} + (y – q)^{2} =r^{2}

Where, p = -2, q = 3,r = 4

Now by substitutingthe values in the above equation, we get

(x – p)^{2} +(y – q)^{2} = r^{2}

(x – (-2))^{2} +(y – 3)^{2} = 4^{2}

(x + 2)^{2} +(y – 3)^{2} = 16

x^{2} +4x + 4 + y^{2} – 6y + 9 = 16

x^{2} + y^{2} +4x – 6y – 3 = 0

∴ The equation of thecircle is x^{2} + y^{2} + 4x – 6y – 3 = 0

(ii) Centre (a, b) andradius

.

Given:

The radius is

and the centre (a, b)

By using the formula,

The equation of thecircle with centre (p, q) and radius ‘r’ is (x – p)^{2} + (y – q)^{2} =r^{2}

Where, p = a, q = b, r=

Now by substitutingthe values in the above equation, we get

(x – p)^{2} +(y – q)^{2} = r^{2}

(x – a)^{2} +(y – b)^{2} =

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

x^{2} + y^{2} –2ax – 2by = 0

∴ The equation of thecircle is x^{2} + y^{2} – 2ax – 2by = 0

(iii) Centre (0, -1) andradius 1.

Given:

The radius is 1 andthe centre (0, -1)

By using the formula,

The equation of thecircle with centre (p, q) and radius ‘r’ is (x – p)^{2} + (y – q)^{2} =r^{2}

Where, p = 0, q = -1,r = 1

Now by substitutingthe values in the above equation, we get

(x – p)^{2} +(y – q)^{2} = r^{2}

(x – 0)^{2} +(y – (-1))^{2} = 1^{2}

(x – 0)^{2} +(y + 1)^{2} = 1

x^{2} + y^{2} +2y + 1 = 1

x^{2} + y^{2} +2y = 0

∴ The equation of thecircle is x^{2} + y^{2} + 2y = 0.

(iv) Centre (a cos α, a sinα) and radius a.

Given:

The radius is ‘a’ andthe centre (a cos α, a sin α)

By using the formula,

The equation of thecircle with centre (p, q) and radius ‘r’ is (x – p)^{2} + (y – q)^{2} =r^{2}

Where, p = a cos α, q= a sin α, r = a

Now by substitutingthe values in the above equation, we get

(x – p)^{2} +(y – q)^{2} = r^{2}

(x – a cosα)^{2} +(y – a sinα)^{2} = a^{2}

x^{2} –(2acosα)x + a^{2}cos^{2}α + y^{2} – (2asinα)y + a^{2}sin^{2}α= a^{2}

We know that sin^{2}θ+ cos^{2}θ = 1

So,

x^{2} –(2acosα)x + y^{2} – 2asinαy + a^{2} = a^{2}

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

∴ The equation of thecircle is x^{2} + y^{2} – (2acosα) x – (2asinα) y =0.

(v) Centre (a, a) andradius √2 a.

Given:

The radius is√2 a and the centre (a, a)

By using the formula,

The equation of thecircle with centre (p, q) and radius ‘r’ is (x – p)^{2} + (y – q)^{2} =r^{2}

Where, p = a, q = a, r= √2 a

Now by substitutingthe values in the above equation, we get

(x – p)^{2} +(y – q)^{2} = r^{2}

(x – a)^{2} +(y – a)^{2} = (√2 a)^{2}

x^{2} –2ax + a^{2} + y^{2} – 2ay + a^{2} = 2a^{2}

x^{2} + y^{2} –2ax – 2ay = 0

∴ The equation of thecircle is x^{2} + y^{2} – 2ax – 2ay = 0.

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

(i) (x – 1)^{2} + y^{2} = 4

(ii) (x + 5)^{2} + (y + 1)^{2} = 9

(iii) x^{2} + y^{2} – 4x + 6y = 5

(iv) x^{2} + y^{2} – x + 2y – 3 = 0

**Answer
2** :

(i) (x – 1)^{2} +y^{2} = 4

Given:

The equation (x – 1)^{2} +y^{2} = 4

We need to find thecentre and the radius.

By using the standardequation formula,

(x – a)^{2} +(y – b)^{2} = r^{2} …. (1)

Now let us convertgiven circle’s equation into the standard form.

(x – 1)^{2} +y^{2} = 4

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

By comparing equation(2) with (1), we get

Centre = (1, 0) andradius = 2

∴ The centre of thecircle is (1, 0) and the radius is 2.

(ii) (x + 5)^{2} +(y + 1)^{2} = 9

Given:

The equation (x + 5)^{2} +(y + 1)^{2} = 9

We need to find thecentre and the radius.

By using the standardequation formula,

(x – a)^{2} +(y – b)^{2} = r^{2} …. (1)

Now let us convertgiven circle’s equation into the standard form.

(x + 5)^{2} +(y + 1)^{2} = 9

(x – (-5))^{2} +(y – ( – 1))^{2} = 3^{2} …. (2)

By comparing equation(2) with (1), we get

Centre = (-5, -1) andradius = 3

∴ The centre of thecircle is (-5, -1) and the radius is 3.

(iii) x^{2} + y^{2} –4x + 6y = 5

Given:

The equation x^{2} +y^{2} – 4x + 6y = 5

We need to find thecentre and the radius.

By using the standardequation formula,

(x – a)^{2} +(y – b)^{2} = r^{2} …. (1)

Now let us convertgiven circle’s equation into the standard form.

x^{2} + y^{2} –4x + 6y = 5

(x^{2} –4x + 4) + (y^{2} + 6y + 9) = 5 + 4 + 9

(x – 2)^{2} +(y + 3)^{2} = 18

(x – 2)^{2} +(y – (-3))^{2} = (3√2)^{2} … (2)

By comparing equation(2) with (1), we get

Centre = (2, -3) andradius = 3√2

∴ The centre of thecircle is (2, -3) and the radius is 3√2.

(iv) x^{2} + y^{2} –x + 2y – 3 = 0

Given:

The equation x^{2} +y^{2} – x + 2y – 3 = 0

We need to find thecentre and the radius.

By using the standardequation formula,

(x – a)^{2} +(y – b)^{2} = r^{2} …. (1)

Now let us convertgiven circle’s equation into the standard form.

x^{2} + y^{2} –x + 2y – 3 = 0

(x^{2} –x + ¼) + (y^{2} + 2y + 1) – 3 – ¼ – 1 = 0

(x – ½)^{2} +(y + 1)^{2} = 17/4 …. (2)

By comparing equation(2) with (1), we get

Centre = (½, – 1) andradius = √17/2

∴ The centre of thecircle is (½, -1) and the radius is √17/2.

**Answer
3** :

Given:

Centre is (1, 2) andwhich passes through the point (4, 6).

Where, p = 1, q = 2

We need to find theequation of the circle.

By using the formula,

(x – p)^{2} +(y – q)^{2} = r^{2}

(x – 1)^{2} +(y – 2)^{2} = r^{2}

It passes through thepoint (4, 6)

(4 – 1)^{2} +(6 – 2)^{2} = r^{2}

3^{2} + 4^{2} =r^{2}

9 + 16 = r^{2}

25 = r^{2}

r = √25

= 5

So r = 5 units

We know that the equationof the circle with centre (p, q) and having radius ‘r’ is given by: (x – p)^{2} +(y – q)^{2} = r^{2}

By substitute thevalues in the above equation, we get

(x – 1)^{2} +(y – 2)^{2} = 5^{2}

x^{2} –2x + 1 + y^{2} – 4y + 4 = 25

x^{2} + y^{2} –2x – 4y – 20 = 0.

∴ The equation of thecircle is x^{2} + y^{2} – 2x – 4y – 20 = 0.

**Answer
4** :

Let us find the pointsof intersection of the lines.

On solving the lines x+ 3y = 0 and 2x – 7y = 0, we get the point of intersection to be (0, 0)

On solving the lines x+ y + 1 and x – 2y + 4 = 0, we get the point of intersection to be (-2, 1)

We have circle withcentre (-2, 1) and passing through the point (0, 0).

We know that theradius of the circle is the distance between the centre and any point on theradius. So, we find the radius of the circle.

So, the equation is (x– p)^{2} + (y – q)^{2} = r^{2}

Where, p = -2, q = 1

(x + 2)^{2} +(y – 1)^{2} = r^{2} …. (1)

Equation (1) passesthrough (0, 0)

So, (0 + 2)^{2} +(0 – 1)^{2} = r^{2}

4 + 1 = a^{2}

5 = r^{2}

r = √5

We know that theequation of the circle with centre (p, q) and having radius ‘r’ is given by: (x– p)^{2} + (y – q)^{2} = r^{2}

By substitute thevalues in the above equation, we get

(x – (-2))^{2} +(y – 1)^{2} = (√5)^{2}

(x + 2)^{2} +(y – 1)^{2} = 5

x^{2} +4x + 4 + y^{2} – 2y + 1 = 5

x^{2} + y^{2} +4x – 2y = 0

∴ The equation of thecircle is x^{2} + y^{2} + 4x – 2y = 0.

**Answer
5** :

It is given that thecentre lies on the positive y – axis at a distance of 6 from the origin, we getthe centre (0, 6).

We have a circle withcentre (0, 6) and having radius 4.

We know that theequation of the circle with centre (p, q) and having radius ‘r’ is given by: (x– p)^{2} + (y – q)^{2} = r^{2}

Where, p = 0, q = 6, r= 4

Now by substitutingthe values in the equation, we get

(x – 0)^{2} +(y – 6)^{2} = 4^{2}

x^{2} + y^{2} –12y + 36 = 16

x^{2} + y^{2} –12y + 20 = 0.

∴ The equation of thecircle is x^{2} + y^{2} – 12y + 20 = 0.

**Answer
6** :

It is given that thecircle has the radius 10 and has diameters 2x + y = 6 and 3x + 2y = 4.

We know that thecentre is the intersection point of the diameters.

On solving thediameters, we get the centre to be (8, -10).

We have a circle withcentre (8, -10) and having radius 10.

By using the formula,

We know that theequation of the circle with centre (p, q) and having radius ‘r’ is given by: (x– p)^{2} + (y – q)^{2} = r^{2}

Where, p = 8, q = -10,r = 10

Now by substitutingthe values in the equation, we get

(x – 8)^{2} +(y – (-10))^{2} = 10^{2}

(x – 8)^{2} +(y + 10)^{2} = 100

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

x^{2} + y^{2} –16x + 20y + 64 = 0.

∴ The equation of thecircle is x^{2} + y^{2} – 16x + 20y + 64 = 0.

Find the equation of the circle

(i) which touches both the axes at a distance of 6 units from the origin.

(ii) Which touches x – axis at a distance of 5 from the origin and radius 6 units.

(iii) Which touches both the axes and passes through the point (2, 1).

(iv) Passing through the origin, radius 17 and ordinate of the centre is – 15.

**Answer
7** :

(i) which touches both theaxes at a distance of 6 units from the origin.

A circle touchesthe axes at the points (±6, 0) and (0, ±6).

So, a circle has acentre (±6, ±6) and passes through the point (0, 6).

We know that theradius of the circle is the distance between the centre and any point on theradius. So, we find the radius of the circle.

So, the equation is (x– p)^{2} + (y – q)^{2} = r^{2}

Where, p = 6, q = 6

(x – 6)^{2} +(y – 6)^{2} = r^{2} …. (1)

Equation (1) passesthrough (0, 6)

So, (0 – 6)^{2} +(6 – 6)^{2} = r^{2}

36 + 0 = r^{2}

r = √36

= 6

^{2} + (y – q)^{2} = r^{2}

Now by substitutingthe values in the equation, we get

(x ± 6)^{2} +(y ± 6)^{2} = (6)^{2}

x^{2 }±12x + 36 + y^{2 }± 12y + 36 = 36

x^{2} + y^{2 }±12x ± 12y + 36 = 0

∴ The equation of thecircle is x^{2} + y^{2 }± 12x ± 12y + 36 = 0.

(ii) Which touches x – axisat a distance of 5 from the origin and radius 6 units.

A circle touchesthe x – axis at the points (±5, 0).

Let us assume thecentre of the circle is (±5, a).

We have a circle withcentre (5, a) and passing through the point (5, 0) and having radius 6.

We know that theradius of the circle is the distance between the centre and any point on theradius. So, we find the radius of the circle.

So, the equation is (x– p)^{2} + (y – q)^{2} = r^{2}

Where, p = 5, q = a

(x – 5)^{2} +(y – a)^{2} = r^{2} …. (1)

Equation (1) passesthrough (5, 0)

So, (5 – 5)^{2} +(0 – 6)^{2} = r^{2}

0 + 36 = r^{2}

r = √36

= 6

We have got the centreat (±5, ±6) and having radius 6 units.

^{2} + (y – q)^{2} = r^{2}

Now by substitutingthe values in the equation, we get

(x ± 5)^{2} +(y ± 6)^{2} = (6)^{2}

x^{2} ±10x + 25 + y^{2} ± 12y + 36 = 36

x^{2} + y^{2} ±10x ± 12y + 25 = 0.

∴ The equation of thecircle is x^{2} + y^{2} ± 10x ± 12y + 25 = 0.

(iii) Which touches both theaxes and passes through the point (2, 1).

Let us assume thecircle touches the x-axis at the point (a, 0) and y-axis at the point (0, a).

Then the centre of thecircle is (a, a) and radius is a.

Its equation will be(x – p)^{2} + (y – q)^{2} = r^{2}

By substituting thevalues we get

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

So now, equation (1)passes through P (2, 1)

By substituting thevalues we get

(2 – a)^{2} +(1 – a)^{2} = a^{2}

4 – 4a + a^{2} +1 – 2a + a^{2} = a^{2}

5 – 6a + a^{2} =0

(a – 5) (a – 1) = 0

So, a = 5 or 1

Case (i)

We have got the centreat (5, 5) and having radius 5 units.

^{2} + (y – q)^{2} = r^{2}

Now by substitutingthe values in the equation we get

(x – 5)^{2} +(y – 5)^{2} = 5^{2}

x^{2} –10x + 25 + y^{2} – 10y + 25 = 25

x^{2} + y^{2} –10x – 10y + 25 = 0.

∴ The equation of thecircle is x^{2} + y^{2} – 10x – 10y + 25 = 0.

Case (ii)

We have got the centreat (1, 1) and having a radius 1 unit.

^{2} + (y – q)^{2} = r^{2}

Now by substitutingthe values in the equation we get

(x – 1)^{2} +(y – 1)^{2} = 1^{2}

x^{2} –2x + 1 + y^{2} – 2y + 1 = 1

x^{2} + y^{2} –2x – 2y + 1 = 0

∴ The equation ofthe circle is x^{2} + y^{2} – 2x – 2y + 1 = 0.

(iv) Passing through theorigin, radius 17 and ordinate of the centre is – 15.

Let us assume theabscissa as ‘a’

We have a circle withcentre (a, – 15) and passing through the point (0, 0) and having radius 17.

By using the distanceformula,

17^{2} =a^{2} + (-15)^{2}

289 = a^{2} +225

a^{2} =64

|a| = √64

|a| = 8

a = ±8 …. (1)

We have got the centreat (±8, – 15) and having radius 17 units.

^{2} + (y – q)^{2} = r^{2}

Now by substitutingthe values in the equation, we get

(x ± 8)^{2} +(y – 15)^{2} = 17^{2}

x^{2 }±16x + 64 + y^{2} – 30y + 225 = 289

x^{2} + y^{2 }±16x – 30y = 0.

∴ The equation of thecircle is x^{2} + y^{2 }± 16x – 30y = 0.

**Answer
8** :

It is given that weneed to find the equation of the circle with centre (3, 4) and touches thestraight line 5x + 12y – 1 = 0.

We have a circle withcentre (3, 4) and having a radius 62/13.

^{2} + (y – q)^{2} = r^{2}

Now by substitutingthe values in the equation, we get

(x – 3)^{2} +(y – 4)^{2} = (62/13)^{2}

x^{2} –6x + 9 + y^{2} – 8y + 16 = 3844/169

169x^{2} +169y^{2} – 1014x – 1352y + 4225 = 3844

169x^{2} +169y^{2} – 1014x – 1352y + 381 = 0

∴ The equation of thecircle is 169x^{2} + 169y^{2} – 1014x – 1352y + 381 =0.

**Answer
9** :

Let us assume thecircle touches the axes at (a, 0) and (0, a) and we get the radius to be |a|.

We get the centre ofthe circle as (a, a). This point lies on the line x – 2y = 3

a – 2(a) = 3

-a = 3

a = – 3

Centre = (a, a) = (-3,-3) and radius of the circle(r) = |-3| = 3

We have circle withcentre (-3, -3) and having radius 3.

^{2} + (y – q)^{2} = r^{2}

Now by substitutingthe values in the equation, we get

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

(x + 3)^{2} +(y + 3)^{2} = 9

x^{2} +6x + 9 + y^{2} + 6y + 9 = 9

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

∴ The equation of thecircle is x^{2} + y^{2} + 6x + 6y + 9 = 0.

**Answer
10** :
It is given that the circle has the centre at the intersection point ofthe lines 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 and passes through the origin

^{2} + (y – q)^{2} = r^{2}

Now by substitutingthe values in the equation, we get

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