# Circle

## Short Questions

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

_{1}, y

_{1}) and B(8, 0) = (x

_{2}, y

_{2}) be the given end points of diameter.

Now, the equation of circle is given by,
(x - x_{1})(x - x_{2}) + (y - y_{1})(y - y_{2}) = 0
or, (x - 0)(x - 8) + (y - 6)(y - 0) = 0

or, x^{2} - 8x + y^{2} - 6y = 0

^{2}+ y

^{2}- 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.

_{1}, y

_{1}) and B(0, 4) = (x

_{2}, y

_{2}) be the given end points of diameter.

Now, the equation of circle is given by,
(x - x_{1})(x - x_{2}) + (y - y_{1})(y - y_{2}) = 0
or, (x - 3)(x - 0) + (y - 0)(y - 4) = 0

or, x^{2} - 3x + y^{2} - 4y = 0

^{2}+ y

^{2}- 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.

_{1}, y

_{1}) and B(2, -5) = (x

_{2}, y

_{2}) be the given end points of diameter.

Now, the equation of circle is given by,
(x - x_{1})(x - x_{2}) + (y - y_{1})(y - y_{2}) = 0
or, (x - 1)(x - 2) + (y - 3)(y + 5) = 0

or, x^{2} - 2x - x + 2 + y^{2} + 5y - 3y - 15 = 0

or, x^{2} + y^{2} - 3x + 2y - 13 = 0

^{2}+ y

^{2}- 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')

Find the center and radius of a circle having the equation x^{2} + y^{2} - 10x - 4y = 7.

Given equation of circle: x

^{2}+ y

^{2}- 10x - 4y = 7

Comparing it with, x^{2} + y^{2} + 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

If the radius of the circle x^{2} + y^{2} - 4x - 6y - k = 0 is 4 units, find the value of *k*.

^{2}+ y

^{2}- 4x - 6y - k = 0

Comparing it with, x^{2} + y^{2} + 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

*k*is 3.

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

_{1}, y

_{1}) and B(6, 5) = (x

_{2}, y

_{2}) is the given end points of diameter.

Now, the equation of circle is given by,
(x - x_{1})(x - x_{2}) + (y - y_{1})(y - y_{2}) = 0
or, (x - 4)(x - 6) + (y - 1)(y - 5) = 0

or, x^{2} - 6x - 4x + 24 + y^{2} - 5y - y + 5 = 0

or, x^{2} + y^{2} - 10x - 6x + 29 = 0

^{2}+ y

^{2}- 10x - 6x + 29 = 0 is the required equation of circle.

If the centre of circle x^{2} + y^{2} - ax - by - 12 = 0 is (2, 3), find the value of
a and b.

^{2}+ y

^{2}- ax - by - 12 = 0

Comparing it with, x^{2} + y^{2} + 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

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

_{1}, y

_{1}) and B(3, 4) = (x

_{2}, y

_{2}) is the given end points of diameter.

Now, the equation of circle is given by,
(x - x_{1})(x - x_{2}) + (y - y_{1})(y - y_{2}) = 0
or, (x - 5)(x - 3) + (y - 6)(y - 4) = 0

or, x^{2} - 3x - 5x + y^{2} - 4y - 6y + 24 = 0

or, x^{2} + y^{2} - 8x - 10y + 39 = 0

^{2}+ y

^{2}- 8x - 10y + 39 = 0 is the required equation of circle.

Find the value of *m* when the centre of a circle with equation x^{2} + y^{2} - mx + 6y - 12 = 0 is (-2, -3).

^{2}+ y

^{2}- mx + 6y - 12 = 0

Comparing it with, x^{2} + y^{2} + 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.

radius(r) = diameter/2 = $\sqrt5$ units

Now, equation of the circle is given by,
(x - h)^{2} + (y - k)^{2} = r^{2}
or, (x - 1)^{2} + (y + 2)^{2} = 5

or, x^{2} - 2x + 1 + y^{2} + 4y + 4 = 5

so, x^{2} + y^{2} - 2x + 4y = 0

^{2}+ y

^{2}- 2x + 4y = 0 is the required equation of circle.

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

radius(r) = 5 units

Now, equation of the circle is given by,
(x - h)^{2} + (y - k)^{2} = r^{2}
or, (x - 2)^{2} + (y - 3)^{2} = 25

or, x^{2} - 4x + 4 + y^{2} - 6y + 9 = 25

so, x^{2} + y^{2} - 4x - 6y - 12 = 0

Hence, x^{2} + y^{2} - 4x - 6y - 12 = 0 is the required equation of circle.

Find the co-ordinates of the centre of a circle having an equation x^{2} + y^{2} + 4x - 5 = 0.

^{2}+ y

^{2}+ 4x - 5 = 0

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

g = 2, f = 0

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

_{1}, y

_{1}) and B(2, -7) = (x

_{2}, y

_{2}) is the given end points of diameter.

Now, the equation of circle is given by,
(x - x_{1})(x - x_{2}) + (y - y_{1})(y - y_{2}) = 0
or, (x - 3)(x - 4) + (y - 2)(y + 7) = 0

or, x^{2} - 2x - 3x + 6 + y^{2} + 7y - 4y - 28 = 0

or, x^{2} + y^{2} - 5x + 3y - 22 = 0

^{2}+ y

^{2}- 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.

radius(r) = 3 units

Now, equation of the circle is given by,
(x - h)^{2} + (y - k)^{2} = r^{2}
or, (x + 1)^{2} + (y + 3)^{2} = 9

or, x^{2} + 2x + 1 + y^{2} + 6y + 9 = 9

so, x^{2} + y^{2} + 2x + 6y + 1 = 0

^{2}+ y

^{2}+ 2x + 6y + 1 = 0 is the required equation of circle.

Find the center of the circle whose equation is x^{2} + y^{2} - 6x + 2y + 1 = 0.

^{2}+ y

^{2}- 6x + 2y + 1 = 0

Comparing it with, x^{2} + y^{2} + 2gx + 2fy + c = 0, we get,
g = 2-3, f = 1