Question

The equation of the circle passing through the point (2,1) and touching y – axis at the origin is
A . ${{x}^{2}}+{{y}^{2}}-5x=0$
B. $2{{x}^{2}}+2{{y}^{2}}-5x=0$
C. ${{x}^{2}}+{{y}^{2}}+5x=0$
D. ${{x}^{2}}-{{y}^{2}}-5x=0$

Answer 1

Hint: In this question, we have to find the equation of the circle passing through the point (2,1) touching on the y- axis touching the origin. As the circle touches the y- axis at the origin, this means the centre will be at x- axis . Then, we find the equation of the circle using the standard form of 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 circle touching the y- axis at the origin
This means k = 0 and r = r units
Hence the equation of the circle is
${{(x-h)}^{2}}+{{(y-0)}^{2}}={{r}^{2}}$
Thus Equation be ${{(x-h)}^{2}}+{{y}^{2}}={{r}^{2}}$------------------- (1)
It is also given that the circle passes through (2,1)
Equation be ${{(2-h)}^{2}}+{{(1)}^{2}}={{r}^{2}}$
That is ${{(2-h)}^{2}}+1={{r}^{2}}$----------------- (2)
It is also passing through origin i.e. (0,0)
Equation be ${{(0-h)}^{2}}+{{0}^{2}}={{r}^{2}}$
This means ${{h}^{2}}={{r}^{2}}$ --------------------- (3)
Now equation (2) becomes
${{(2-h)}^{2}}+1={{h}^{2}}$
That is $4-4h+{{h}^{2}}+1={{h}^{2}}$
$-4h+5 = 0$
That is h = $\dfrac{5}{4}$
Now, from equation (1) and (3), we get equation of circle is
${{(x-h)}^{2}}+{{y}^{2}}={{h}^{2}}$
${{x}^{2}}-2xh+{{y}^{2}}=0$
Now we put the value of h, we get
${{x}^{2}}-2x\left( \dfrac{5}{4} \right)+{{y}^{2}}=0$
$4{{x}^{2}}-10x+4{{y}^{2}}=0$
By taking 2 common from the above equation, we get
That is $2{{x}^{2}}-5x+2{{y}^{2}}=0$
That is $2{{x}^{2}}-5x+2{{y}^{2}}=0$ is the required equation of the circle.

Thus, Option (B ) is correct.

Note: We know standard form of circle is ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$
where the centre is (h,k) and radius is r.
If the circle touches the y-axis at origin, then the centre will be at x- axis.
So the centre = (h,0) and radius = h