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CAT General Equation of Circle - Practice Questions & MCQ

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

Syllabus

Quick Facts

5 Questions around this concept.

Solve by difficulty

EASY

The equation of the circle centred at the origin and having a radius of 6 units is:

$x^2+y^2=6$

$x^2+y^2=30$

$x^2+y^2+36=0$

$x^2+y^2=36$

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Concepts Covered - 1

General Equation of Circle

Now, let's find the equation of of circle

Let P(x, y) be any point on the circumference of the circle with centre at C (h,k).

Then, by definition, | CP | = r. It means that the distance from the point C to point P is constant, which is radius r.

Using the distance formula, we have

$\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\sqrt{(x-h)^{2}+(y-k)^{2}}=r}\\\\\mathrm{i.e.\;\;\;\;\;\;\;\;\;\;\;(x-h)^{2}+(y-k)^{2}=r^{2}}$

If the centre of the circle is the origin or (0,0) (i.e h =0 and k = 0)

Then the equation of the circle becomes

$\\\mathrm{\;\;\;\;\;\;\;\left (x-0 \right )^2+\left (y-0 \right )^2=r^2}\\\\\Rightarrow \mathrm{\quad\quad\quad\quad\;\;\;\;\;\; x^2+y^2=r^2}$

In the previous concept, we learn that the equation of circle when center, C(h,k) and radius r given is $\mathrm{ (x-h)^2+(y-k)^2=r^2}$ .

Now simplify this using the identity $\mathrm{ \left ( a-b \right )^2=a^2-2ab+b^2}$

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

Compare the above equation with $\mathrm{x^2+y^2+2gx+2fy+c=0 }$

we get,

$\begin{aligned} \mathrm{h}&=\mathrm{-g}\\\mathrm{k}&=\mathrm{-f}\\\mathrm{h^2+k^2-r^2}&=\mathrm{c}\\\text{or }\mathrm{h^2+k^2-c}&=\mathrm{r^2} \end{aligned}$

Therefore the equation of any circle can be expressed in the form

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

This is known as the general equation of the circle.

Again,

$\begin{aligned} &x^{2}+y^{2}+2 g x+2 f y+c=0\\ &\begin{array}{l} \Rightarrow\left(x^{2}+2 g x+g^{2}\right)+\left(y^{2}+2 f y+f^{2}\right)=g^{2}+f^{2}-c \\ \Rightarrow(x+g)^{2}+(y+f)^{2}=(\sqrt{g^{2}+f^{2}-c})^{2} \\ \Rightarrow\{x-(-g)\}^{2}+\{y-(-f)\}^{2}=(\sqrt{g^{2}+f^{2}-c})^{2} \end{array} \end{aligned}$

This is of the form $\mathrm{(x-h)^{2}+(y-k)^{2}=r^{2}}$ which represents a circle having centre at (- g, -f) and radius $\mathrm{\sqrt{g^{2}+f^{2}-c}}$ .