Circle : Equation of Circle , Equation of Circle in different cases , Position of a Point with respect to circle.

Equation of Circle

Let C(i, j) be the centre of the circle and CP(r) be the radius of the Circle, then the equation of the circle is

Circle equation

${(x-i)}^{2}+{(y-j)}^{2}={r}^{2}\, -----(1)$

Now, origin (0, 0) be the centre of the circle then equation 1 becomes,

${x}^{2}+{y}^{2}={r}^{2}$

The area of the circle is given by

$\pi {r}^{2}\, sq.\, unit$

General Equation of Circle

The General equation of the second degree may represent a circle if the coeffcient of ${x}^{2}$ and coeffcient of ${y}^{2}$ are identical and coeffcient of xy becomes zero i.e

${ax}^{2}+{by}^{2}+2ixy\, +2gx+2fy\, +c\, =0\, ----(1)$

represents a circle if a =b i.e

$coefficient\, of\, {x}^{2}\, =\, coefficient\, {y}^{2}$

$coefficient\, of\, xy\, =0\, ,$

then equation 1 reduces as

${x}^{2}+{y}^{2}+2gx+2fy\, +\, c\, =0$

whose centre and radius are

$(-g,\, -f)\, \, and\, \sqrt {{g}^{2}+{f}^{2}-c}\, \, \, respectively$

Equation of circle in diameter form

Let

$A({x}_{1},{y}_{1})\, and\, B({x}_{2},{y}_{2})\,$ be the end points of a diameter of the given circle and let P(x,y) be any point on the circle such that

$<APB={90}^{0}$

Slope of AP

${m}_{1}=\frac {y-{y}_{1}} {x-{x}_{1}}\, and\, slope\, of\, BP\, ,\, {m}_{2}=\frac {y-{y}_{2}} {x-{x}_{2}}$

for a perpendicular, m1 * m2 = 1 i.e AP * BP = -1

$\frac {y-{y}_{1}} {x-{x}_{1}}\, *\frac {y-{y}_{2}} {x-{x}_{2}}=-1$

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

Equation of Circle in different cases

Case 1:

When circle passes through origin (0,0). let the equation of the circle be

${(x-i)}^{2}+{(y-j)}^{2}={r}^{2}\, -----(1)$

it passes through origin (0,0)

${i}^{2}+{j}^{2}\, ={r}^{2}$

equation 1 becomes

${(x-i)}^{2}+{(y-j)}^{2}={i}^{2}+{j}^{2}$

${x}^{2}+{y}^{2}-2ix-2jy\, =0$

Case 2:

When circle passes through x- axis. let the centre of the circle be C(i, j) and toches x-axis at point P , then radius of the circle is

$CP\, =\, \left | {j} \right |$

equation of the circle

${(x-i)}^{2}+{(y-j)}^{2}={CP}^{2}={j}^{2}$

${x}^{2}+{y}^{2}-2ix-2jy+{i}^{2}\, =0$

Case 3:

When circle passes through y- axis. let the centre of the circle be C(i, j) and toches y-axis at point P , then radius of the circle is

$CP\, =\, \left | {i} \right |$

equation of the circle

${(x-i)}^{2}+{(y-j)}^{2}={CP}^{2}={i}^{2}$

${x}^{2}+{y}^{2}-2ix-2jy+{j}^{2}\, =0$

Case 4:

When circle toches both axis. In this case

$\left | {i} \right |=\left | {j} \right |=\alpha$

then equation of the circle

${(x-i)}^{2}+{(y-j)}^{2}={r}^{2}\, \, where\, \left | {i} \right |=\left | {j} \right |=\left | {r} \right |=\alpha$

${x}^{2}+{y}^{2}\pm 2\alpha x\pm 2\alpha y+{\alpha }^{2}=0$

Post By : Rahul Kumar 29 Jan, 2020 2829 views Maths

Circle : Equation of Circle , Equation of Circle in different cases , Position of a Point with respect to circle.

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