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**Equation of a Circle**

**Equation of a circle**is a second degree equation in

**and**

*x***which satisfies all the points on the circumference of a circle in a Cartesian Plane. If**

*y***is the length of radius and**

*r***is the centre of a circle, then this circle has an equation**

*(a, b)***.**

*(x – a)*^{2}+ (y – b)^{2}= r^{2}An equation of the form

**represents a circle if**

*x*^{2}+ y^{2}+ 2gx + 2fy + c = 0**, and is then an equation of the circle with centre at**

*g*^{2}+ f^{2}– c > 0**and radius**

*(–g, –f)**.*

**√(g**^{2}**+ f**^{2}**– c)**###
**Equation of a Circle With Centre at
Origin**

Let O(0, 0) be the centre of a circle. If P(x, y) be any
point on the circumference of the circle, then OP = r.

By using distance formula,

OP

^{2}= (x – 0)^{2}+ (y – 0)^{2}
∴ x

^{2}+ y^{2}= r^{2}… … … (i)
The relation (i) is satisfied by every point on the circle,
hence it represents the equation of circle with centre at origin and radius r.

*Example 1:*

*Find the equation of circle with centre at origin and radius 5 units.*

*Solution:*

*Here, radius*(r) = 5 units

*The equation of a circle with centre at origin is*

*x*

^{2}+ y^{2}= r^{2}*or, x*

^{2}+ y^{2}= 5^{2}

*∴*

*x*

^{2}+ y^{2}= 25 is the required equation.###
**Equation of Circle With Centre at (h, k)**

Let C(h,
k) be the centre and r be the radius of a circle. If P(x, y) be any point on
the circle, then CP = r.

By using
distance formula,

CP

^{2}= (x – h)^{2}+ (y – k)^{2}
or, (x – h)

^{2}+ (y – k)^{2}= r^{2}… … … (i)
This
relation is true for every point on the circle. Hence, represents a circle with
centre at (h, k) and radius r.

Equation
(i) can be written as

x

^{2}–2hx+h^{2}+y^{2}–2ky+k^{2}–r^{2}= 0
or, x

^{2}+y^{2}–2hx–2ky+h^{2}+k^{2}–r^{2}= 0
or, x

^{2}+ y^{2}+ 2gx + 2fy + c = 0 … … … (ii)
where g = -h, f = -k and c = h

^{2}+ k^{2}– r^{2}.
The
equation (ii) is said to be the

*General equation of circle*. This equation of circle has the following characteristics.
a.
It
is second degree in x and y.

b.
Coefficient
of x

^{2}= coefficient of y^{2}
c.
Coefficient
of xy = 0.

*Thus the general equation of second degree ax*

^{2}+ 2hxy + by^{2}+ 2gx + 2fy + c = 0 will represent a circle if a = b and h = 0.

*Example 2:**Find the equation of a circle whose centre is at point (-4, 3) and radius 4 units.*

**Solution:**

*Here, centre of circle is (-4, 3)*

*i.e. h = -4, k = 3 and radius (r) = 4.*

*The equation of circle with centre:(h, k) and radius=r is*

*(x–h)*

^{2}+(y – k)^{2}= r^{2}*or, (x + 4)*

^{2}+ (y – 3)^{2}= 4^{2}*or, x*

^{2}+ 8x +16 + y^{2}– 6y + 9 = 16*∴*

*x*

^{2}+ y^{2}+ 8x – 6y + 9 = 0 is the required equation.

*Example 3:**Find the centre and radius of the circle x*^{2}+ y^{2}– 6x + 4y – 23 = 0

**Solution:**

*The given equation is*

*x*

^{2}+y^{2}–6x+4y–23 = 0*or, x*

^{2}– 6x + y^{2}+ 4y – 23 = 0*or, x*

^{2}– 2.x.3 + 3^{2}+ y^{2}+ 2.y.2 + 2^{2}= 23 + 3^{2}+ 2^{2}*or, (x – 3)*

^{2}+ (y + 2)^{2}= 23 + 9 + 4*or, (x – 3)*

^{2}+ (y + 2)^{2}= 6^{2}*Comparing it with (x – h)*

^{2}+ (y – k)^{2}= r^{2}, we have h = 3, k = -2 and r = 6*Hence, centre of circle is (h, k) = (3, - 2) and radius r = 6.*

###
**Equation
of Circle in a Diameter Form**

Let O be
the centre and A(x

_{1}, y_{1}) and B(x_{2}, y_{2}) be the ends of diameter AB of the circle. If P(x, y) be any point on the circumference of the circle, then
Since,
∠APB is the angle in
the semi-circle ∠APB = 90°.

∴ m

_{1}× m_{2}= -1
or, (y – y

_{1})(y – y_{2}) = - (x –x_{1})(x – x_{2})
or, (x – x

_{1})(x – x_{2}) + (y – y_{1})(y – y_{2}) = 0
Hence,
it is the equation of circle in diameter form.

*Example 4:**Find the equation of a circle whose ends of the diameter are at (-1, 3) and (5, 7).*

**Solution:**

*Here, ends of the diameter of the circle are (-1, 3) and (5, 7).*

*i.e. x*

_{1}= -1, y_{1}= 3, x_{2}= 5 and y_{2}= 7.*The equation of circle in diameter form is*

*(x–x*

_{1})(x–x_{2}) + (y–y_{1})(y–y_{2}) = 0*or, (x + 1)(x – 5) + (y – 3)(y – 7) = 0*

*or, x*

^{2}– 5x + x – 5 + y^{2}– 7y – 3y + 21 = 0*or, x*

^{2}+ y^{2}– 4x – 10y + 16 = 0 is the required equation.

*Example 5:**Find the equation of a circle with centre (-3, 6) and touching the x-axis.*

**Solution:**

*Then the equation of the circle is*

*(x–h)*

^{2}+ (y–k)^{2}= r^{2}*or, (x + 3)*

^{2}+ (y – 6)^{2}= 6^{2}*or, x*

^{2}+ 6x + 9 + y^{2}– 12y + 36 = 36

*∴*

*x*

^{2}+y^{2}+6x -12y + 9 = 0 is the required equation or circle.

*You can comment your questions or problems regarding the equation of circle here.*

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