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CAT Different Form of the Equation of the Circle - Practice Questions & MCQ

Edited By admin | Updated on Oct 04, 2023 04:20 PM | #CAT

Syllabus

Quick Facts

4 Questions around this concept.

Solve by difficulty

EASY

Which of the following circles passing through the origin is:

$x^2+y^2+4x=0$

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

$x^2+y^2=4$

both (a) and (b)

View Solution

Concepts Covered - 1

Different Form of the Equation of the Circle

1. When a circle passes through origin

The equation of a circle with centre at (h, k) and radius equal to a is $\mathrm{(x-h)^2+(y-k)^2=a^2}$ .

As circle passes through the origin (0,0), put x = 0 and y = 0

We get, $\mathrm{a^{2}=h^{2}+k^{2}}$

Therefore, the equation of the circle $\mathrm{(x-h)^2+(y-k)^2=a^2}$ becomes

$\\\Rightarrow \mathrm{(x-h)^2+(y-k)^2=h^{2}+k^{2}}\\\Rightarrow \mathrm{x^{2}+y^{2}-2 h x-2 k y=0}$

Also, we know that the general equation of the circle is $\mathrm{x^2+y^2+2gx+2fy+c=0 }$

As circle passes through the origin (0,0), put x = 0 and y = 0

We get, c = 0

So, the equation of a circle passing through the origin is

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

With, centre at (-g, -f) and radius = $\mathrm{ \sqrt{g^2+f^2} }$ .

2. When a circle touches x-axis

The equation of a circle with centre at (h, k) and radius equal to a is $\mathrm{(x-h)^2+(y-k)^2=a^2}$ .

If a circle touches the x-axis, then the y-coordinate of the centre will be equal to the radius of the circle.

Hence, the equation of the circle $\mathrm{ (x-h)^2+(y-k)^2=a^2}$ becomes

$\\\Rightarrow\mathrm{ (x-h)^2+(y-k)^2=k^2}\\\Rightarrow\mathrm{ x^{2}+y^{2}-2 h x-2 a y+h^2=0}$

3. When a circle touches y-axis

The equation of a circle with centre at (h, k) and radius equal to a is $\mathrm{(x-h)^2+(y-k)^2=a^2}$ .

If a circle touches the y-axis, then the x-coordinate of the centre will be equal to the radius of the circle. Hence, the equation of the circle $\mathrm{(x-h)^2+(y-k)^2=a^2}$ becomes

$\\\Rightarrow \mathrm{(x-h)^2+(y-k)^2=h^2}\\\Rightarrow \mathrm{x^{2}+y^{2}-2 a x-2 k y+k^2=0}$

5. When a circle touches both x-axis and y-axis

The equation of a circle with centre at (h, k) and radius equal to a is $\mathrm{(x-h)^2+(y-k)^2=a^2}$ .

If a circle touches both the coordinate axes then the abscissa as well as ordinate of the centre will be equal to the radius of the circle. Hence, the equation of the circle will be of the form:

$\\\Rightarrow \mathrm{(x-a)^{2}+(y-a)^{2}=a^{2}}\\ \Rightarrow \mathrm{x^{2}+y^{2}-2 a x-2 a y+a^{2}=0}$