# Finding Equation of Circle in Two Dimensional Geometry

## Definition of Circle

If a point is moving in a plane at equal distances from a fixed point, its path of transmission or locus is called a circle. The fixed point is called the center of the circle and the fixed distance from the moving fixed point is called the radius of the circle.

### Some Formulae on Circle

1. The equation of a circle having center at origin and radius ‘a’ is {{x}^{2}}+{{y}^{2}}={{a}^{2}}.........\left( 1 \right)
The parametric equation of the circle (1) is

$x=a\cos \theta ,\,\,y=a\sin \theta$

2. The equation of a circle with center at (α, β) and radius ‘a’ is

${{\left( x-\alpha \right)}^{2}}+{{\left( y-\beta \right)}^{2}}={{a}^{2}}$

3. The general equation of a circle is {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0 with center at \left( -g,-f \right) and radius = \sqrt{{{g}^{2}}+{{f}^{2}}-c}.
The circle makes x-intercept of 2\sqrt{{{g}^{2}}-c} unit and y-intercept of 2\sqrt{{{f}^{2}}-c} unit.

4. The quadratic equation a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0 will represent the equation of a circle if a = b and h = 0.

5. If (x1, y1) and (x2, y2) be the end points of diameter of circle then the equation of the circle is $\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)+\left( y-{{y}_{1}} \right)\left( y-{{y}_{2}} \right)=0$

6. Equation of the circle passing through the intersecting points of the circles {{x}^{2}}+{{y}^{2}}+2{{g}_{1}}x+2{{f}_{1}}y+{{c}_{1}}=0 and {{x}^{2}}+{{y}^{2}}+2{{g}_{1}}x+2{{f}_{1}}y+c\,=\,0, {{x}^{2}}+{{y}^{2}}+2{{g}_{2}}x+2{{f}_{2}}y+{{c}_{2}}=0 is ${{x}^{2}}+{{y}^{2}}+2{{g}_{1}}x+2{{f}_{1}}y+{{c}_{1}}+k\left( {{x}^{2}}+{{y}^{2}}+2{{g}_{2}}x+2{{f}_{2}}y+{{c}_{2}} \right)=0\,\,\left[ k\ne 0 \right]$

7. Equation of the concentric circle with the circle {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0 is  ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c’=0$

8. The point (x1, y1) will be in, on or outer of the circle {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0 if ${{x}^{2}}+{{y}^{2}}+2{{g}_{1}}x+2{{f}_{1}}y+c>0,\,\,=\,\,or\,\,<0$

9. Equation of the common chord of the circles {{x}^{2}}+{{y}^{2}}+2{{g}_{1}}x+2{{f}_{1}}y+c\,=\,0 and {{x}^{2}}+{{y}^{2}}+2{{g}_{1}}x+2{{f}_{1}}y+c\,=\,0 is $2\left( {{g}_{1}}-{{g}_{2}} \right)x+2\left( {{f}_{1}}-{{f}_{2}} \right)y+{{c}_{1}}-{{c}_{2}}=0$

 Example 01

A circle passes through the point (5, –1) and its center is at (2, –3). Find the equation of the circle.

Solution:

The circle passes through the point (5, –1) and its center is at (2, –3). So its radius = \sqrt{{{\left( 5-2 \right)}^{2}}+{{\left( -1+3 \right)}^{2}}}=\sqrt{9+4}=\sqrt{13}
Therefore the equation of the circle is

${{\left( x-2 \right)}^{2}}+{{\left( y+3 \right)}^{2}}={{\left( \sqrt{13} \right)}^{2}}$

$\Rightarrow {{x}^{2}}-4x+4+{{y}^{2}}+6y+9=13$

$\Rightarrow {{x}^{2}}+{{y}^{2}}-4x+6y=0$

 Example 02

Find the center and radius of the circle 5{{x}^{2}}+5{{y}^{2}}-8x+6y-15=0.

Solution:

We have,

$5{{x}^{2}}+5{{y}^{2}}-8x+6y-15=0$

$\Rightarrow {{x}^{2}}+{{y}^{2}}-\frac{8}{5}x+\frac{6}{5}y-3=0$

Comparing it with the general form of the circle {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0, we get

$2g=-\frac{8}{5};\,\,2f=\frac{6}{5}\,\,and\,\,c=-3$

$\Rightarrow g=-\frac{4}{5};\,\,f=\frac{3}{5}\,\,and\,\,c=-3$

So, the center of the circle is \left( -g,-f \right)=\left( \frac{4}{5},-\frac{3}{5} \right) and radius = \sqrt{{{g}^{2}}+{{f}^{2}}-c}=\sqrt{{{\left( -\frac{4}{5} \right)}^{2}}+{{\left( \frac{3}{5} \right)}^{2}}-\left( -3 \right)}=\sqrt{4}=2 units

 Example 03

Find the parametric equation of the circle {{x}^{2}}+{{y}^{2}}-5x+2y+5=0.

Solution:

We have,

${{x}^{2}}+{{y}^{2}}-5x+2y+5=0$

$\Rightarrow {{x}^{2}}-2\times x\times \frac{5}{2}+{{\left( \frac{5}{2} \right)}^{2}}+{{y}^{2}}+2\times y\times 1+{{\left( 1 \right)}^{2}}={{\left( \frac{5}{2} \right)}^{2}}+{{\left( 1 \right)}^{2}}-5$

$\Rightarrow {{\left( x-\frac{5}{2} \right)}^{2}}+{{\left( y+1 \right)}^{2}}=\frac{9}{4}$

$\Rightarrow {{\left( x-\frac{5}{2} \right)}^{2}}+{{\left( y+1 \right)}^{2}}={{\left( \frac{3}{2} \right)}^{2}}$

Obviously, x-\frac{5}{2}=\frac{3}{2}\cos \theta \,\,and\,\,y+1=\frac{3}{2}\sin \theta satisfies the above equation.
Therefore, the parametric equation of the given circle is

$x-\frac{5}{2}=\frac{3}{2}\cos \theta \,\,and\,\,y+1=\frac{3}{2}\sin \theta$

$\Rightarrow x=\frac{1}{2}\left( 5+3\cos \theta \right)\,\,and\,\,y=-1+\frac{3}{2}\sin \theta$

 Example 04

If 2{{x}^{2}}+2{{y}^{2}}+3x-y-5=0 be the inner circle of an equilateral triangle, then find the area of the triangle.

Solution:

We have

$2{{x}^{2}}+2{{y}^{2}}+3x-y-5=0$

$\Rightarrow {{x}^{2}}+{{y}^{2}}+\frac{3}{2}x-\frac{1}{2}y-\frac{5}{2}=0$

Therefore radius of the above circle r = \sqrt{{{\left( \frac{3}{4} \right)}^{2}}+{{\left( -\frac{1}{4} \right)}^{2}}-\left( -\frac{5}{2} \right)}=\sqrt{\frac{9}{16}+\frac{1}{16}+\frac{5}{2}}=\frac{1}{4}\sqrt{50} units
Now, let each side of the equilateral triangle be ‘a’ and area be △, then

$\Delta =\frac{\sqrt{3}}{4}{{a}^{2}}\,\,and\,\,\Delta =rs=r\times \frac{3a}{2}\,\,\left[ \because s=Half\,of\,perimeter=\frac{3a}{2} \right]$

$\therefore \frac{\sqrt{3}}{4}{{a}^{2}}=r\times \frac{3a}{2}$

$\Rightarrow a=2\sqrt{3}r$

$\therefore \Delta =\frac{\sqrt{3}}{4}{{a}^{2}}=\frac{\sqrt{3}}{4}{{\left( 2\sqrt{3}r \right)}^{2}}=\frac{\sqrt{3}}{4}\times 12{{r}^{2}}$

$=3\sqrt{3}\times \frac{50}{16}\,\,\left[ \because \,r=\frac{1}{4}\sqrt{50} \right]$

$=\frac{75\sqrt{3}}{8}$

Therefore, area of the equilateral triangle is \frac{75\sqrt{3}}{8} sq. units.

 Example 05

The center of a circle lies on the straight line 3x-2y+8=0 and passes through the points (2, 3) and (–6, –1). Find the equation of the circle.

Solution:

Let the equation of the circle be

${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0………\left( 1 \right)$

Since, (1) passes through the points (2, 3) and (–6, –1)

${{\left( 2 \right)}^{2}}+{{\left( 3 \right)}^{2}}+2g\left( 2 \right)+2f\left( 3 \right)+c=0$

$\Rightarrow 4+9+4g+6f+c=0$

$\Rightarrow 4g+6f+c+13=0………\left( 2 \right)$

${{\left( -6 \right)}^{2}}+{{\left( -1 \right)}^{2}}+2g\left( -6 \right)+2f\left( -1 \right)+c=0$

$\Rightarrow 36+1-12g-2f+c=0$

$\Rightarrow -12g-2ff+c+37=0………\left( 3 \right)$

Again the center \left( -g,-f \right) of the given circle lies on the straight line 3x-2y+8=0

$\therefore 3\left( -g \right)-2\left( -f \right)+8=0$

$\Rightarrow -3g+2f+8=0……….\left( 4 \right)$

Now subtracting (3) from (2) we get,

$16g+8f-24=0$

$\Rightarrow 2g+f-3=0………….\left( 5 \right)$

Solving (4) and (5) we get, g=2,\,\,f=-1
Putting the values of g\,and\,f in (2) we get, c = –15
Therefore, required equation of the circle is

${{x}^{2}}+{{y}^{2}}+4x-2y-15=0$

 Example 06

If a circle touches the x-axis at the origin and passes through the point (h, k), find the equation of the circle.

Solution:

Since the circle touches the x-axis at the origin, so the ordinate of the center of the circle is equal to the radius of the circle.
Let the radius of the circle = a and coordinate of the center is (α, a). Then the equation of the circle will be

${{\left( x-\alpha \right)}^{2}}+{{\left( y-a \right)}^{2}}={{a}^{2}}………\left( 1 \right)$

According the question circle (1) passes through (0, 0) and (h, k)

${{\left( 0-\alpha \right)}^{2}}+{{\left( 0-a \right)}^{2}}={{a}^{2}}$

$\Rightarrow {{\alpha }^{2}}+{{a}^{2}}={{a}^{2}}\Rightarrow \alpha =0$

${{\left( h-\alpha \right)}^{2}}+{{\left( k-a \right)}^{2}}={{a}^{2}}$

$\Rightarrow {{h}^{2}}-2\alpha h+{{\alpha }^{2}}+{{k}^{2}}-2ak+{{a}^{2}}={{a}^{2}}$

$\Rightarrow {{h}^{2}}+{{k}^{2}}-2ak+{{a}^{2}}={{a}^{2}}\,\,\left[ \because \,a=0 \right]$

$\therefore \,\,2a=\frac{{{h}^{2}}+{{k}^{2}}}{k}$

Therefore, equation of the circle is

${{x}^{2}}+{{y}^{2}}-2ay=0\,\,\left[ Putting\,\alpha =0\,\,in\,\left( 1 \right) \right]$

$\Rightarrow {{x}^{2}}+{{y}^{2}}-\frac{{{h}^{2}}+{{k}^{2}}}{k}\times y=0\,\left[ \because \,\,2a=\frac{{{h}^{2}}+{{k}^{2}}}{k} \right]$

$\therefore k\left( {{x}^{2}}+{{y}^{2}} \right)=\left( {{h}^{2}}+{{k}^{2}} \right)y$

 Example 07

Find the equation of the circle passing through the point of intersection of the circle {{x}^{2}}+{{y}^{2}}+4\left( x+y \right)+4=0 and the straight linex+y+2=0, also the center at the origin.

Solution:

${{x}^{2}}+{{y}^{2}}+4\left( x+y \right)+4=0………\left( 1 \right)$

$x+y+2=0………\left( 2 \right)$

Equation of the circle passing through the point of intersection of (1) and (2) is

${{x}^{2}}+{{y}^{2}}+4\left( x+y \right)+4+k\left( x+y+2 \right)=0$

$\Rightarrow {{x}^{2}}+{{y}^{2}}+\left( 4+k \right)x+\left( 4+k \right)y+4+2k=0……….\left( 3 \right)$

Now, the coordinate of the circle (3) is \left( -\frac{4+k}{2},-\frac{4+k}{2} \right)
According to the question

$-\frac{4+k}{2}=0$

$\Rightarrow 4+k=0\Rightarrow k=-4$

Therefore, equation of the required circle is

${{x}^{2}}+{{y}^{2}}-4=0$

 Example 08

Show that the circles {{x}^{2}}+{{y}^{2}}+6x+14y+9=0 and {{x}^{2}}+{{y}^{2}}-4x-10y-7=0 touches each other externally. Find the common tangent of both the circles.

Solution:

${{x}^{2}}+{{y}^{2}}+6x+14y+9=0………\left( 1 \right)$

${{x}^{2}}+{{y}^{2}}-4x-10y-7=0………\left( 2 \right)$

Comparing circle (1) with the general form of the circle {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0, we have the center of the circle (1) \left( -g,-f \right)=A\left( -3,-7 \right) and radius = {{r}_{1}}=\sqrt{{{g}^{2}}+{{f}^{2}}-c}=\sqrt{9+49-9}=7

Similarly, comparing circle (2) with the general form of the circle {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0, we have the center of the circle (2) \left( -g,-f \right)=B\left( 2,5 \right) and {{r}_{1}}=\sqrt{{{g}^{2}}+{{f}^{2}}-c}=\sqrt{4+25-\left( -7 \right)}=6.

Now, the distance between the centers of both the circles

$\overline{AB}=\sqrt{{{\left( -3-2 \right)}^{2}}+{{\left( -7-5 \right)}^{2}}}=\sqrt{25+144}=\sqrt{169}=13$

\overline{AB}=7+6={{r}_{1}}+{{r}_{2}}= Sum of the radius of the circles.
Therefore, the circles touch each other externally.
Now the equation of any curve passing through point of contact is

${{x}^{2}}+{{y}^{2}}+6x+14y+9+k\left( {{x}^{2}}+{{y}^{2}}-4x-10y-7 \right)=0$

$\Rightarrow \left( 1+k \right)\left( {{x}^{2}}+{{y}^{2}} \right)+\left( 6-4k \right)x+\left( 14-10k \right)y+9-7k=0………\left( 3 \right)$

Equation (3) will represent a straight line if 1+k=0\Rightarrow k=-1
Putting k=-1 in (3) we have

$10x+24y+16=0$

$\Rightarrow 5x+12y+8=0$

Hence the equation of the common tangent of both the circles.

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