Question

The equation of the circle with center (4, 3) and touching the circle ${{x}^{2}}+{{y}^{2}}=1$ is:
(a) ${{x}^{2}}+{{y}^{2}}-8x-6y+9=0$
(b) ${{x}^{2}}+{{y}^{2}}+8x+6y-11=0$
(c) ${{x}^{2}}+{{y}^{2}}-8x-6y-11=0$
(d) ${{x}^{2}}+{{y}^{2}}+8x+6y-9=0$

Answer 1

Hint: To solve this question, we should know about property when two circles touch each other, that is, for two circles with center at ${{C}_{1}}$ and ${{C}_{2}}$ of radii ${{r}_{1}}$ and ${{r}_{2}}$ touches each other if ${{C}_{1}}{{C}_{2}}={{r}_{1}}\pm {{r}_{2}}$. Also, we should know that for the circle ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ center is (– g, – f) and radius is $\sqrt{{{g}^{2}}+{{f}^{2}}-c}$.

Complete step-by-step answer:
In this question, we have to find the equation of the circle which has a center at (4, 3) and touches ${{x}^{2}}+{{y}^{2}}=1$. Let us consider the equation of circle C as ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$. We have been given that (4, 3) is the center of the circle ${{C}_{1}}$ and we know that the center of the circle is given by (– g, – f). Therefore, we can say g = – 4 and f = – 3. So, we can write the equation of the circle as ${{x}^{2}}+{{y}^{2}}-2\left( 4 \right)x-2\left( 3 \right)y+c=0$.
${{x}^{2}}+{{y}^{2}}-8x-6y+c=0.....\left( i \right)$
We can also say that the center of the circle $C'={{x}^{2}}+{{y}^{2}}=1$ is (0, 0) and radius = r’ = 1. We have been given that the circle C and C’ touch each other and we know that when two circles touch each other then the distance between their centers is equal to the algebraic sum of their radii, that is,
$CC'=r\pm r'.....\left( ii \right)$
Now, we know the value of r’ = 1 and C (4, 3) and C’ (0, 0). So, by using distance formula, we can write,
$CC'=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$
$CC'=\sqrt{16+9}$
$CC'=\sqrt{25}$
$CC'=5$
And, we know that the radius of the circle is given by $\sqrt{{{g}^{2}}+{{f}^{2}}-c}$ when we have been given the equation of the circle. So, we can write the radius of circle C as
$\sqrt{{{\left( 4 \right)}^{2}}+{{\left( 3 \right)}^{2}}-c}$
$=\sqrt{16+9-c}$
$=\sqrt{25-c}$
Now, we will put all these values in the equation (ii). So, we will get,
$5=1\pm \sqrt{25-c}$
$\Rightarrow 5-1=\pm \sqrt{25-c}$
Now, we will square both sides of the equation. So, we will get,
${{\left( 4 \right)}^{2}}={{\left( \pm \sqrt{25-c} \right)}^{2}}$
$16=25-c$
$c=25-16$
$c=9$
Now, we will put the value of c in equation (i). So, we will get,
${{x}^{2}}+{{y}^{2}}-8x-6y+9=0$
This is the equation of the circle with center (4, 3) and touches ${{x}^{2}}+{{y}^{2}}=1$.
Therefore, option (a) is the right answer.

Note: The possible mistake one can make is by writing the center (4, 3) in the equation as ${{x}^{2}}+{{y}^{2}}+8x+6y+c=0$ which is wrong because the center of the equation ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ is (– g, – f), we should substitute $g=-4$ and $f=-3$. Also, students make mistakes by calculating the radius with the formula $\sqrt{{{g}^{2}}+{{f}^{2}}+c}$ which is wrong. The correct formula for the radius is $\sqrt{{{g}^{2}}+{{f}^{2}}-c}$.