Question

Equation of the circle concentric with the circle ${{x}^{2}}+{{y}^{2}}-6x+12y+15=0$and of double its area is
A . ${{x}^{2}}+{{y}^{2}}-3x+12y-15=0$
B. ${{x}^{2}}+{{y}^{2}}-3x+12y-30=0$
C. ${{x}^{2}}+{{y}^{2}}-6x+12y-25=0$
D. ${{x}^{2}}+{{y}^{2}}-6x+12y-20=0$

Answer 1

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Hint: In this question, we have to find the equation of the circle which is concentric with the given equation. For this we find out the radius and then area. As the circles are concentric, this means the circle has two different radii with common centre, so we find the centre and radii of the given equation by using the standard form of circle and then we use that points of centre in another equation to get the another radii so that we are able to get our required equation.
Formula Used:
${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$
Where $(h,k)$is the centre of circle.
Where $h=-g\,and\,\,k=-f$
Complete Step – by – Step Solution:
We have given an equation ${{x}^{2}}+{{y}^{2}}-6x+12y+15=0$ -------------- (1)
We have to find the equation of circle concentric with the given circle.
As the equation ${{x}^{2}}+{{y}^{2}}-6x+12y+15=0$
Now we solve the equation by completing the squares, we get
$({{x}^{2}}-6x+9)+({{y}^{2}}+12y+36)+15-9-36=0$
${{(x-3)}^{2}}+{{(y+6)}^{2}}-45+15=0$
${{(x-3)}^{2}}+{{(y+6)}^{2}}=30$
 Then we get r = $\sqrt{30}$
Now, we know area of circle =$\pi {{r}^{2}}$= 30 $\pi $square units
Area of required circle = 2 ( 30$\pi $) = $\pi {{r}^{2}}$
That is r = $\sqrt{60}$
Radius of required circle is $\sqrt{60}$units
As both of the circles are concentric
Therefore, Centre of required circle = Centre of given circle
Now we find out the centre of the given circle
We know standard form of circle is ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$
Now we compare the equation (1) with the standard form of circle, we get
$2g=-6\,\,and\,\,2f=12\,\,and\,\,c=15$
We get $g=-3\,\,and\,\,f=6\,and\,\,c=15$[1]
We know centre of circle is (-g,-f) = (3, - 6)
Hence the equation of required circle is
${{(x-3)}^{2}}+{{(y+6)}^{2}}={{\left( \sqrt{60} \right)}^{2}}$
Hence, the equation of required circle is ${{x}^{2}}+{{y}^{2}}-3x+12y-15=0$
Thus, Option (A) is correct.
Note :- In these types of questions, students made mistakes in finding the values and putting the equations from one equation to another. By doing a lot of practice, there will be less chances of confusion and we get the correct answer.