(i) Equation of circle with centre (0, 0) and radius r (ii) Equation of circle with centre (h, k) and radius r

**Equation
of a circle in standard form**

Let the centre be *C*(0, 0) and radius be *r *and *P*(*x*,
*y*) be the moving point.

Note that the point *P *having coordinates (*x*, *y*)
is represented as *P*(*x*, *y*) .

Then, *CP *= *r *and so *CP*^{2}
= *r*^{2}

Therfore (*x *− 0)^{2} + ( *y *− 0)^{2}
= *r*^{2}

That is *x*^{2} + *y*^{2} = *r*^{2}

This is the equation of the circle with centre (0, 0) and radius r

Let the centre be *C*(*h*, *k *) and *r *be
the radius and *P*(*x*, *y*) be the moving point,

Then, *CP *= *r and *so *CP*^{2} = *r*^{2}
.

That is, (*x *− *h*)^{2} + ( *y *− *k *)^{2}
= *r*^{2} .This is the equation of the circle in Standard form,
which is also known as centre-radius form.

Expanding the equation, we get

*x*^{2} + *y*^{2} − 2*hx *− 2*ky *+ *h*^{2}
+ *k*^{2} − *r*^{2} = 0 .

Taking 2*g *= −2*h*, 2 *f *= −2*k*, *c *=
*h*^{2} + *k*^{2} − *r*^{2} , the
equation takes the form

*x*^{2} + *y*^{2} + 2*gx *+ 2 *fy *+ *c *= 0
, called the general form of a circle.

The equation *x*^{2} + *y*^{2} + 2*gx *+
2 *fy *+ *c *= 0 is a second degree equation in *x *and *y *possessing
the following characteristics:

(i) It is a second degree equation in *x *and
*y *,

(ii) coefficient of *x*^{2} =
coefficient of *y*^{2} ≠ 0 ,

(iii) coefficient of *xy *= 0 .

Conversely, we prove that an equation possessing these three
characteristics, always represents a circle. Let* *

*ax*^{2} + *ay*^{2} + 2*g*′*x *+ 2 *f
y*′ + *c *= 0 ………………………………(1)

be a second degree equation in *x *and *y *having
characteristics (i), (ii) and (iii), *a* ≠ 0 . Dividing (1) by a, gives

equation (2) becomes *x*^{2} + *y*^{2} + 2*gx *+ 2 *fy *+ *c
*= 0 .

Adding and subtracting *g*^{2} and *f*^{2} , we get *x*^{2}
+ 2*gx *+ *g*^{2} + *y*^{2}
+ 2 *fy *+ *f *^{2} - *g*^{2}
- *f *^{2} + *c *= 0

⇒ (*x *+ *g*)^{2} + ( *y *+ *f
*)^{2} = *g *^{2} + *f *^{2} – *c*

⇒ (*x *- (-*g*))^{2}
+ ( *y *- (- *f *))^{2} = ( √[ *g*^{2} + *f *^{2} - *c*]* *)^{2}

This is in the standard
form of a circle with
centre C (-*g*, -
*f *) and
radius *r *= √[ *g*^{2} + *f *^{2} - *c*].
Hence equation (1) represents a circle with centre (-*g*, - *f *) = ( -*g’*/a , - *f’/a* ) and
radius

**Note**

The equation
*x*^{2}
+ *y*^{2}
+ 2*gx** *+ 2 *fy *+ *c *= 0 represents a circle
with centre (–*g*, –*f** *)
and radiu √[ *g*^{2}
+
*f *^{2}
- *c*].

(i) It is a real circle if
*g*^{2} + *f*^{2} - *c *> 0 ;

(ii) a point circle if *g*^{2} + *f *^{2} - *c *= 0 ;

(iii) or an imaginary
circle if *g*^{2} + *f *^{2} - *c *< 0 with no locus.

**Example 5.1**

Find the general equation of a circle with centre (−3, −4) and
radius 3 units.

**Solution**

Equation of the circle in standard form is
( *x *- *h*)^{2} + ( *y *- *k *)^{2}
= *r *^{2}

⇒ (*x *- (-3))^{2}
+ ( *y *- (-4))^{2}
= 3^{2}

⇒ ( *x *+ 3)^{2}
+ ( *y *+ 4)^{2} = 32

⇒ *x*^{2} + *y*^{2}
+ 6*x *+ 8 *y *+16 = 0 .

**Theorem 5.1**

The circle passing through the points of intersection (real or
imaginary) of the line *lx *+ *my *+ *n *= 0 and the circle *x*^{2} + *y*^{2}
+ 2*gx *+ 2 *fy *+ *c *= 0 is the circle of the form

*x*^{2} + *y*^{2} + 2*gx *+ 2 *fy *+ *c *+ *λ
*(*lx *+ *my *+ *n*) = 0 , *λ *∈ **R**^{1}.

Proof

Let the circle be S : *x*2 + *y*2 + 2*gx *+ 2 *fy
*+ = 0 , … (1)

and the line be L : lx + my + n = 0 . … (2)

Consider S + λL = 0
. That is *x*^{2} + *y*^{2}
+ 2*gx* + 2*fy* + c +λ*(lx + my + n)* =
0
... (3)

Grouping the terms of *x, y* and
constants, we get

x^{2} + y^{2} + *x*(2g +λ*l*)+ y(2*f* +λ*m*)+ c +λ*n* = 0 which is a second degree equation in *x* and *y* with coefficients
of x^{2} and y^{2} equal and there is no *xy* term.

If (α, β) is a point of intersection of S and L satisfying
equation (1) and (2), then it satisfies equation (3).

Hence S + λL = 0 represents the required circle.

**Example 5.2**

Find the equation of the circle described on the chord 3*x *+
*y *+ 5 = 0 of the circle *x*^{2} + *y*^{2} = 16
as diameter.

**Solution**

Equation of the circle passing through the points of
intersection of the chord and circle by Theorem 5.1 is *x*^{2} + *y*^{2}
-16 + *l** *(3*x *+ *y *+
5) = 0 .

The chord 3*x *+ *y *+ 5 = 0 is a diameter of this
circle if the centre lies on the chord.

Therefore, the equation of the required circle is *x*^{2}
+ *y*^{2} + 3*x *+ *y *-11 = 0.

**Example 5.3**

Determine whether *x *+ *y *−1 = 0 is the equation of
a diameter of the circle
*x*^{2} + *y*^{2}
− 6*x *+ 4 *y *+ *c *= 0 for all possible values of *c *.

**Solution**

Centre of the circle is (3, -2) which lies on *x *+ *y *-1
= 0 . So the line *x *+ *y *-1 = 0 passes through the centre and
therefore the line *x *+ *y *-1 = 0 is a diameter of the circle for
all possible values of *c *.

**Theorem 5.2**

The equation of a circle with (*x*_{1} , *y*_{1}
) and (*x*_{2} , *y*_{2} ) as extremities of one of
the diameters of the
circle is (*x *− *x*_{1}
)(*x *− *x*_{2} ) + ( *y *− *y*_{1})( *y *−
*y*_{2}) = 0.

**Proof**

Let *A*(*x*_{1} , *y*_{1} ) and *B*(*x*_{2}
, *y*_{2} ) be the two extremities of the diameter *AB *, and
*P*(*x*, *y*) be any point on the circle. Then ÐAPB = π/2 (angle in a semi-circle).

Therefore, the product of slopes of *AP *and *PB *is
equal to -1.

That is, = −1 yielding the equation of the required circle as

(*x *- *x*_{1}
)(*x *- *x*_{2} ) + ( *y *- *y*_{1} )( *y
*- *y*_{2} ) = 0 .

**Example 5.4**

Find the general equation of the circle whose diameter is the line
segment joining the points (−4, −2) and (1,1) .

**Solution**

Equation of the circle with end points of the diameter as( *x*_{1}
, *y*_{1} ) and ( *x*_{2} , *y*_{2} )
given in

theorem 5.2 is

( *x *- *x*_{1} )( *x *- *x*_{2}
) + ( *y *- *y*_{1} )( *y *- *y*_{2} ) = 0

⇒ ( *x *+ 4)( *x *-1) + ( *y *+
2)( *y *-1) = 0

⇒
*x*^{2} + *y*^{2}
+ 3*x *+ *y *- 6 = 0 which is the required equation of the circle.

**Theorem 5.3**

The position of a point *P*(*x*_{1} , *y*_{1}
) with respect to a given circle *x*^{2} + *y*^{2} + 2*gx *+ 2 *fy *+ *c *= 0 in the Plane containing the circle is Outside
or on or inside the circle according as *x*^{2} + *y*^{2} + 2*gx *+ 2 *fy *+ *c *is

**Proof**

Equation of the circle is *x*^{2} + *y*^{2}
+ 2*gx *+ 2 *fy *+ *c *= 0 with centre C (-*g*, - *f *)
and radius *r *=

Let *P*(*x*_{1}, *y*_{1} ) be a
point in the plane. Join *CP *and let it meet the circle at *Q *.Then
the point *P *is outside, on or within the circle according as

**Example 5.5**

Examine the position of the point (2, 3) with respect to the
circle *x*^{2} + *y*^{2} − 6*x *− 8 *y *+12
= 0.

**Solution**

Taking (*x*_{1}, *y*_{1}) as (2, 3),
we get

*x*^{2} + *y*^{2} + 2*gx _{1} *+ 2

= 4 + 9 -12 - 24 + 12 = -11 < 0.

Therefore, the point (2, 3) lies inside the circle, by theorem 5.3.

**Example 5.6**

The line 3*x *+ 4 *y *−12 = 0 meets the coordinate
axes at *A *and *B *. Find the equation of the circle drawn on *AB *as
diameter.

**Solution**

Writing the line 3*x *+ 4*y *= 12, in intercept form
yields *x/4 *+ *y/3 *= 1. Hence the points *A *and *B *are (4, 0) and(0, 3) .

Equation of the circle in diameter form is

( *x *- *x*_{1} )( *x *- *x*_{2}
) + ( *y *- *y*_{1} )( *y *- *y*_{2} ) = 0

( *x *- 4)( *x *- 0) + ( *y *- 0)( *y *- 3)
= 0

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

**Example 5.7**

A line 3*x *+ 4 *y *+10 = 0 cuts a chord of length 6
units on a circle with centre of the circle (2,1) . Find the equation of the
circle in general form.

**Solution**

*C*(2,1) is the centre and 3*x *+ 4 *y *+10 = 0 cuts a
chord *AB *on the circle. Let *M *be the midpoint of *AB *. Then we
have

AM = BM = 3 . Now BMC is a right triangle.

So, we have CM = = 4.

By Pythogoras theorem *BC*^{2} = *BM*^{2}
+ *MC*^{2} = 3^{2} + 4^{2} = 25

*BC *= 5 = radius.

So, the equation of the required circle is (*x *- 2)^{2}
+ ( *y *-1) = 5^{2}

*x*^{2} + *y*^{2} - 4*x *- 2 *y *– 20 = 0.

**Example 5.8**

A circle of radius 3 units touches both the axes. Find the
equations of all possible circles formed in the general form.

**Solution**

As the circle touches both the axes, the distance of the centre
from both the axes is 3 units, centre can be (±3,
±3) and hence there are four
circles with radius 3,
and the required equations of the four circles are *x*^{2} + *y*^{2} ± 6*x *± 6 *y *+
9 = 0 .

**Example 5.9**

Find the centre and radius of the circle 3*x*^{2} +
(*a *+1) *y*^{2} + 6*x *- 9*y *+ *a *+ 4 = 0.

**Solution**

Coefficient of *x*^{2} = Coefficient of *y*^{2}
(characteristic (ii) for a second degree equation to represent a circle).

That is, 3 = *a *+1 and *a *= 2 .

Therefore, the equation of the circle is

3*x*^{2} + 3*y*^{2} + 6*x *- 9 *y
*+ 6 = 0

*x*^{2} + *y*^{2} + 2*x *- 3*y *+ 2 = 0

So, centre is ( -1, 3/2 ) and radius r =

**Example 5.10**

Find the equation of the circle passing through the points
(1,1), (2, −1) , and (3, 2) .

**Solution**

Let the general equation of the circle be

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

It passes through points (1,1), (2, -1) and (3, 2) .

Therefore,
2*g *+ 2 *f *+
*c = *-2 , … (2)

Therefore, the required equation of the circle is

Þ *x*^{2}
+ *y*^{2} - 5*x *- *y *+ 4 = 0

**Note**

Three points on a circle determine the equation to the circle
uniquely. Conversely three equidistant points from a centre point forms a
circle.

Tags : Formula, Solved Example Problems , 12th Mathematics : UNIT 5 : Two Dimensional Analytical Geometry II

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12th Mathematics : UNIT 5 : Two Dimensional Analytical Geometry II : Equation of a circle in standard form | Formula, Solved Example Problems

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