Question
Question: If a circle \({{x}^{2}}+{{y}^{2}}=9\) touches the circle \({{x}^{2}}+{{y}^{2}}+6y+c=0\), then c is e...
If a circle x2+y2=9 touches the circle x2+y2+6y+c=0, then c is equal to
A. -27B. 36C. -36D. 27
Solution
To solve this question first we compare the given equations of the circle with the general equation of the circle. For finding the solution first we need to calculate the radius and center of the circle given in the question. If x2+y2+2gx+2fy+c=0 is the general equation of the circle then, the center of circle is given by (−g,−f) and radius of circle is given by r=g2+f2−c .
Then, by using the relation between the center of the circle and radius of the circle we calculate the value of c.
Complete step-by-step answer:
We have given that a circle x2+y2=9 touches the circle x2+y2+6y+c=0.
We have to find the value of c .
Now, we know that if x2+y2+2gx+2fy+c=0 is the general equation of the circle then, the center of circle is given by (−g,−f) and radius of circle is given by r=g2+f2−c
Now consider the first circle x2+y2=9
On comparing with the general equation of circle x2+y2+2gx+2fy+c=0, we get
g=0f=0c=−9
So, the center of circle will be C1=(0,0) and the radius will be
r=g2+f2−c
r1=02+02+9r1=9r1=3
Now, let us consider circle x2+y2+6y+c=0
On comparing with the general equation of circle x2+y2+2gx+2fy+c=0, we get
g=0f=3c=c
So, the center of circle will be C2=(0,−3) and the radius will be
r2=g2+f2−c
r2=02+32−cr2=9−c
Now, we know that if two circles touch each other then, the relation between their centers and radius is given by
C1C2=r2−r1
Substituting the values, we get
⇒3=9−c−3⇒3+3=9−c⇒6=9−c⇒36=9−c⇒36−9=−c−c=27c=−27
The value of c is −27 .
So, the correct answer is “Option A”.
Note: The relation between the center and radius of two circles touch each other is also given by C1C2=r1−r2 and C1C2=r1+r2. But when we put the values and solve these equations we get the answer c=9, which is not matches the options given so we use the relation C1C2=r2−r1.
C1C2=r1−r23=3−9−c3−3=−9−c−9−c=0c=9
And