Question

The equation of the circle with center at (1,-2 ) and passing through the center of the given circle ${{x}^{2}}+{{y}^{2}}-2x+4y+3=0$ is
( a ) ${{x}^{2}}+{{y}^{2}}-2x+4y+3=0$
( b ) ${{x}^{2}}-{{y}^{2}}-2x+4y-3=0$
( c ) ${{x}^{2}}+{{y}^{2}}+2x-4y-3=0$
( d ) ${{x}^{2}}+{{y}^{2}}+2x-4y+3=0$

Answer 1

Hint: To solve this question, we use the standard form of circle and by finding out the values of f,c and g by comparing the equation with the standard form of circle and putting the values in the formula of radius, we get our desirable equation.

Complete Step by step solution:
We have given the circle with centre at (1,-2) which passess through the centre of the given circle ${{x}^{2}}+{{y}^{2}}+2y-3=0$--------- (1)
We have to find the equation of the circle.
We know the standard form of circle = ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$--------- (2)
Compare the equation (1) with equation (2), we get
2g = 0 and 2f = 2
Hence,
g = 0 and f = 1
And we get c = -3
We know center = (-g,-f) = ( 0, -1 )
Now we find the equation of circle that has a center of (1,-2) and passess through (0,-1)
We know that the equation of circle is given by
${{(x-a)}^{2}}+{{(y-b)}^{2}}={{r}^{2}}$
(a,b) is the center of circle is (1,-2) and r is the radius of circle.
That is equation of circle = $1{{(0-1)}^{2}}+{{(-1+2)}^{2}}={{r}^{2}}$
That is 1+1 = ${{r}^{2}}$
This means ${{r}^{2}}$ = 2
So, the required equation of circle = ${{(x-1)}^{2}}+{{(y+2)}^{2}}=2$
By expanding the above equation, we get
 ${{x}^{2}}-2x+1+{{y}^{2}}+4y+4=2$
That is ${{x}^{2}}+{{y}^{2}}-2x+4y+3=0$
Thus, Option ( A ) is correct.

Note: Students made mistakes while finding the values and putting the right equations. We must have a lot of practice to solve these types of questions.