Question

The equation of the circle with centre on the x – axis, radius 4 and passing through the origin is
A . ${{x}^{2}}+{{y}^{2}}+4x=0$
B. ${{x}^{2}}+{{y}^{2}}-8y=0$
C. ${{x}^{2}}+{{y}^{2}}+8x=0$
D. ${{x}^{2}}+{{y}^{2}}+8y=0$

Answer 1

Hint: In this question, we have to find the equation of the circle with centre on the x- axis , radius 4 and passing through the origin. As the centre lies on the x- axis then the y-coordinate of the centre will be zero, this means k = 0. Then, we find the equation of the circle and we get the two values of h. Now we find out the two equations with the different values of h and choose out the correct option.

Formula Used:
Equation of circle = ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$

Complete Step- by- Step Solution:
Let the equation of the required circle be
${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$
We are given that the radius of the circle is 4 and its centre lies on the x – axis.
This means k = 0 and r = 4 units
Hence the equation of the circle is
${{(x-h)}^{2}}+{{(y-0)}^{2}}={{4}^{2}}$
Thus Equation be ${{(x-h)}^{2}}+{{y}^{2}}=16$
It is also given that the circle passes through origin i.e. (0,0)
Equation be ${{(0-h)}^{2}}+{{0}^{2}}=16$
This means ${{h}^{2}}=16$ or h = $\pm 4$
Now we solve it in two cases :-
Case 1 :-
When h = -4, then equation of circle is
${{(x+4)}^{2}}+{{y}^{2}}=16$
On expanding the equation, we get
${{x}^{2}}+{{y}^{2}}+8x+16=16$
That is ${{x}^{2}}+{{y}^{2}}+8x=0$ is the required equation of the circle.
Case 2 :-
When h = 4, then the equation of circle is
${{(x-4)}^{2}}+{{y}^{2}}=16$
On expanding the equation, we get
${{x}^{2}}+{{y}^{2}}-8x+16=16$
That is ${{x}^{2}}+{{y}^{2}}-8x=0$ is the required equation of the circle.

Thus, Option ( C ) is correct.

Note: We know the standard form of equation of circle = ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$
If the centre lies on the x – axis
This means the y-coordinate of the centre will be zero.
Then the equation becomes ${{(x-h)}^{2}}+{{y}^{2}}={{r}^{2}}$