Question

If we have two circles $C_1:x^2+y^2-12x-16y=0$ and $C_2:x^2+y^2+12x-16y=0$ then find equations of circles concentric to circles $C_1$ and $C_2$ respectively such that the difference between their radius is $2$ and they touch each other.

• A. $C_1^{\prime }:x^2+y^2-12x-16y+51=0$ and
  $C_2^{\prime }:x^2+y^2+12x-16y+19=0$
• B. $C_1^{\prime }:x^2+y^2-12x-16y+19=0$ and
  $C_2^{\prime }:x^2+y^2+12x-16y+91=0$
• C. $C_1^{\prime }:x^2+y^2-12x-16y+51=0$ and
  $C_2^{\prime }:x^2+y^2+12x-16y+75=0$
• D. $C_1^{\prime }:x^2+y^2-12x-16y+19=0$ and
  $C_2^{\prime }:x^2+y^2+12x-16y+75=0$

Answer 1

The correct option is C $C_1^{\prime }:x^2+y^2-12x-16y+51=0$ and

$C_2^{\prime }:x^2+y^2+12x-16y+75=0$
Equation of circle $x^2+y^2+2gx+2fy+c=0$ has centre $(-g,-f)$

$c_1=x^2+y^2-12x-16y=0$ have centre $O_1=(6,8)$

$c_2=x^2+y^2+12x-16y=0$ have centre $O_2=(-6,8)$

Concentric circles means circles having same centre.

$\therefore$ Equation of circle concentric to $c_1$ is

$c_1^{\prime }:x^2+y^2-12x-16y+a=0$ ...... $\left(i\right)$ having centre $O_1^{\prime }=(6,8)$

Equation of circle concentric to $c_2$ is

$c_2^{\prime }:x^2+y^2+12x-16y+d=0$ ........ $\left(ii\right)$ having centre $O_2^{\prime }(-6,8)$

Now, it is given that diff. between radius of circles $c_1^{\prime }$ and $c_2^{\prime }$ is $2$.

$\therefore \left(\sqrt{{(-6)}^2+{(-8)}^2-a}\right)-\left(\sqrt{6^2+{(-8)}^2-d}\right)=2$

$⟹\left(\sqrt{100-a}\right)-\left(\sqrt{100-d}\right)=2$ .......... $\left(iii\right)$

It is given that $c_1^{\prime }$ and $c_2^{\prime }$ touch each other.

$\therefore Radius\left(c_1^{\prime }\right)+Radius\left(c_2^{\prime }\right)=Distancebetweentheircentres.$

$\therefore \sqrt{100-a}+\sqrt{100-d}=\sqrt{{(6+6)}^2+{(8+8)}^2}$ ....... (From $\left(i\right),\left(ii\right)$)

$(\sqrt{100-a)}+(\sqrt{100-d})=12$ .......... $\left(iv\right)$

Now adding $\left(iii\right)$ and $\left(iv\right)$

$⟹2\sqrt{100-a}=14$

$⟹100-a=49$

$⟹a=51$

Substituting value of $a$ in eq $\left(iv\right)$

$⟹\sqrt{100-51}+\sqrt{100-d}=12$

$⟹100-d=25$

$⟹d=75$

Thus, the equation of required circles is

$c_1^{\prime }:x^2+y^2-12x-16y+51=0$

$c_2^{\prime }:x^2+y^2+12x-16y+75=0$