Question

The equation of circle which passess through the centre of circle ${{x}^{2}}+{{y}^{2}}+8x+10y-7=0$ and is concentric with the circle ${{x}^{2}}+{{y}^{2}}-4x-6y=0$is
( a ) ${{x}^{2}}+{{y}^{2}}+8x+10y+59=0$
( b ) ${{x}^{2}}+{{y}^{2}}+8x+10y-59=0$
( c ) ${{x}^{2}}+{{y}^{2}}-4x-6y+87=0$
( d ) ${{x}^{2}}+{{y}^{2}}-4x-6y-87=0$

Answer 1

Hint: In this question we have given two equations. In this question from the equation ${{x}^{2}}+{{y}^{2}}-4x-6y=0$, we find the center of circle which is in the form of (-g,-f) by comparing it with standard form of circle. As the equation ${{x}^{2}}+{{y}^{2}}+8x+10y-7=0$ passess through center of circle, from there we find the value of c by putting the values in the given equation and after solving it , we get our desired equation.

Complete Step by step solution:
Given equation is ${{x}^{2}}+{{y}^{2}}-4x-6y=0$ --------------- (1)
We know the standard form of circle is ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$------------- (2)
Now, compare the equation (1) with equation (2), we get
2g = -4 and 2f = - 6 and c = 0
Hence, circle with center = (-g,-f) g = - 2, f = - 3 , c = 0
Then (-g,-f ) = (2, 3)
Now equation of circle concentric with ${{x}^{2}}+{{y}^{2}}+8x+10y-7=0$ is
${{x}^{2}}+{{y}^{2}}+8x+10y+c=0$
As it passess through (2,3 )
Then ${{(2)}^{2}}+{{(3)}^{2}}+8(2)+10(3)+c=0$
That is 59 + c =0
Hence, c = -59
Then the equation of circle is ${{x}^{2}}+{{y}^{2}}+8x+10y-59=0$
Thus, Option ( B) is correct.

Note: In these types of questions, students make mistakes in finding from which equation we have to find the value of g and f as they get confused in two equations. They must have the proper knowledge of solving that type of questions. So students have to practice a lot of questions of this type so that they cannot get confused and get the right answer.