# Circle

## Short Questions

The end point of a diameter of a circle are (0, 6) and (8, 0). Find the equation of the circle.

Let A(0, 6) = (x1, y1) and B(8, 0) = (x2, y2) be the given end points of diameter.

Now, the equation of circle is given by, (x - x1)(x - x2) + (y - y1)(y - y2) = 0 or, (x - 0)(x - 8) + (y - 6)(y - 0) = 0
or, x2 - 8x + y2 - 6y = 0

Hence, x2 + y2 - 8x - 6y = 0 is the required equation of circle.

The endpoint of a diameter of a circle are (3, 0) and (0, 4). Find the equation of the circle.

Let A(3, 0) = (x1, y1) and B(0, 4) = (x2, y2) be the given end points of diameter.

Now, the equation of circle is given by, (x - x1)(x - x2) + (y - y1)(y - y2) = 0 or, (x - 3)(x - 0) + (y - 0)(y - 4) = 0
or, x2 - 3x + y2 - 4y = 0

Hence, x2 + y2 - 3x - 4y = 0 is the required equation of circle.

Find equation of the circle having (1, 3) and (2, -5) as the end points of a diameter.

Let A(1, 3) = (x1, y1) and B(2, -5) = (x2, y2) be the given end points of diameter.

Now, the equation of circle is given by, (x - x1)(x - x2) + (y - y1)(y - y2) = 0 or, (x - 1)(x - 2) + (y - 3)(y + 5) = 0
or, x2 - 2x - x + 2 + y2 + 5y - 3y - 15 = 0
or, x2 + y2 - 3x + 2y - 13 = 0

Hence, x2 + y2 - 3x + 2y - 13 = 0 is the required equation of circle.

Find the co-ordinates of the centre of a circle having equations of two diameters x + y = 5 and 2x - y = 1.

Here,
Given equation of diameters of circles are: x + y = 5
or, x = 5 - y --- (i)
And, 2x - y = 1 --- (ii)

Since, the intersection point of 2 diameter, that is, (x, y) is the center of the circle.
Solving (ii) Using (i), 2(5 - y) = 1
or, 10 - 3y = 1
So, y = 3 and x = 2 (from 'i')

Hence, the co-ordinates of the centre of a circle is (x, y) = (2, 3).

Find the center and radius of a circle having the equation x2 + y2 - 10x - 4y = 7.

Here,
Given equation of circle: x2 + y2 - 10x - 4y = 7

Comparing it with, x2 + y2 + 2gx + 2fy + c = 0, we get,
g = -5, f = -2, c = -7

So, center of circle is (h, k) = (-g, -f) = (5, 2).
And, radius of circle is given by,
r = $\sqrt{g^2 + f^2 - c}$ = $\sqrt{\text{25 + 4 - (-7)}}$
= $\sqrt{36}$
= 6 units

Hence, center of the given circle is (5, 2) radius is 6 units.

If the radius of the circle x2 + y2 - 4x - 6y - k = 0 is 4 units, find the value of k.

Given equation of circle: x2 + y2 - 4x - 6y - k = 0

Comparing it with, x2 + y2 + 2gx + 2fy + c = 0, we get,
g = -2, f = -3, c = -k

Since, radius(r) is given as 4 units, we know, r = $\sqrt{g^2 + f^2 - c}$ or, 4 = $\sqrt{\text{4 + 9 + k}}$
Squaring on both sides, we get,
or, 16 = 13 + k
so, k = 3

Hence, the value of k is 3.

Find the equations of a circle having end points of a diameter (4, 1) and (6, 5).

Let A(4, 1) = (x1, y1) and B(6, 5) = (x2, y2) is the given end points of diameter.

Now, the equation of circle is given by, (x - x1)(x - x2) + (y - y1)(y - y2) = 0 or, (x - 4)(x - 6) + (y - 1)(y - 5) = 0
or, x2 - 6x - 4x + 24 + y2 - 5y - y + 5 = 0
or, x2 + y2 - 10x - 6x + 29 = 0

Hence, x2 + y2 - 10x - 6x + 29 = 0 is the required equation of circle.

If the centre of circle x2 + y2 - ax - by - 12 = 0 is (2, 3), find the value of a and b.

Given equation of circle: x2 + y2 - ax - by - 12 = 0

Comparing it with, x2 + y2 + 2gx + 2fy + c = 0, we get,
g = -$\frac a2$, f = -$\frac b2$ Now, we know the center of circle = (h, k) = (-g, -f) = ($\frac a2$, $\frac b2$)

We have given that, (h, k) = (2, 3).
So, Comparing the value, we get,
$\frac a2$ = 2 and $\frac b2$ = 3

Hence, a = 2 and b = 6.

Find the equation of the circle having the end points of a diameter A(5, 6) and B(3, 4).

Let A(5, 6) = (x1, y1) and B(3, 4) = (x2, y2) is the given end points of diameter.

Now, the equation of circle is given by, (x - x1)(x - x2) + (y - y1)(y - y2) = 0 or, (x - 5)(x - 3) + (y - 6)(y - 4) = 0
or, x2 - 3x - 5x + y2 - 4y - 6y + 24 = 0
or, x2 + y2 - 8x - 10y + 39 = 0

Hence, x2 + y2 - 8x - 10y + 39 = 0 is the required equation of circle.

Find the value of m when the centre of a circle with equation x2 + y2 - mx + 6y - 12 = 0 is (-2, -3).

Given equation of circle: x2 + y2 - mx + 6y - 12 = 0

Comparing it with, x2 + y2 + 2gx + 2fy + c = 0, we get,
g = -$\frac m2$, f = 3
Now, we know the center of circle = (h, k) = (-g, -f) = ($\frac m2$, -3)

We have given that, (h, k) = (-2, -3).
So, Comparing the value, we get,
$\frac m2$ = -2 and -3 = -3

Hence, m = -4.

Find the equation of a circle having centre (1, -2) and the diameter 2$\sqrt5$ units.

Here, center(h, k) = (1, -2)
radius(r) = diameter/2 = $\sqrt5$ units

Now, equation of the circle is given by, (x - h)2 + (y - k)2 = r2 or, (x - 1)2 + (y + 2)2 = 5
or, x2 - 2x + 1 + y2 + 4y + 4 = 5
so, x2 + y2 - 2x + 4y = 0

Hence, x2 + y2 - 2x + 4y = 0 is the required equation of circle.

Find the equation of a circle having centre (2, 3) and radius 5 units.

Here, center(h, k) = (2, 3)

Now, equation of the circle is given by, (x - h)2 + (y - k)2 = r2 or, (x - 2)2 + (y - 3)2 = 25
or, x2 - 4x + 4 + y2 - 6y + 9 = 25
so, x2 + y2 - 4x - 6y - 12 = 0

Hence, x2 + y2 - 4x - 6y - 12 = 0 is the required equation of circle.

Find the co-ordinates of the centre of a circle having an equation x2 + y2 + 4x - 5 = 0.

Given equation of circle: x2 + y2 + 4x - 5 = 0

Comparing it with, x2 + y2 + 2gx + 2fy + c = 0, we get,
g = 2, f = 0

Hence, the co-ordinates of the centre of a given circle is (h, k) = (-g, -f) = (-2 , 0).

Find the equation of the circle whose end points of a diameter are A(3, 4) and B(2, -7).

Let, A(3, 4) = (x1, y1) and B(2, -7) = (x2, y2) is the given end points of diameter.

Now, the equation of circle is given by, (x - x1)(x - x2) + (y - y1)(y - y2) = 0 or, (x - 3)(x - 4) + (y - 2)(y + 7) = 0
or, x2 - 2x - 3x + 6 + y2 + 7y - 4y - 28 = 0
or, x2 + y2 - 5x + 3y - 22 = 0

Hence, x2 + y2 - 5x + 3y - 22 = 0 is the required equation of circle.

Find the equation of a circle having the centre (-1, -3) and radius 3 units.

Here, center(h, k) = (-1, -3)

Now, equation of the circle is given by, (x - h)2 + (y - k)2 = r2 or, (x + 1)2 + (y + 3)2 = 9
or, x2 + 2x + 1 + y2 + 6y + 9 = 9
so, x2 + y2 + 2x + 6y + 1 = 0

Hence, x2 + y2 + 2x + 6y + 1 = 0 is the required equation of circle.

Find the center of the circle whose equation is x2 + y2 - 6x + 2y + 1 = 0.

Given equation of circle: x2 + y2 - 6x + 2y + 1 = 0

Comparing it with, x2 + y2 + 2gx + 2fy + c = 0, we get, g = 2-3, f = 1

Hence, the co-ordinates of the centre of a given circle is (h, k) = (-g, -f) = (3 , -1).