Question

For the circles ${{x}^{2}}+{{y}^{2}}-10x+16y+89-{{r}^{2}}=0$ and ${{x}^{2}}+{{y}^{2}}+6x-14y+42=0$ which of the following is/are true
A) Number of integral values of $r$ are $14$ for which circles are intersecting
B) Number of integral values of $r$ are $9$ for which circles are intersecting
C) For $r$ equal to $13$ number of common tangents are $3$
D) For $r$ equal to $21$ number of common tangents are $2$

Answer 1

We are given the equations of two circles. First find their radii and their centres. After that, check whether the sum of their radii is equal/greater/less than the distance between their centres. Note that, if two circles intersect then the sum of their radii is greater than the distance between their centres. Try it, you will get the answer.

Complete step by step solution:
We know that, general equation of the centre is $a{{x}^{2}}+b{{y}^{2}}+2gx+2fy+c=0$.
Its centre is located at $(-g,-f)$.
And its radius is given by $\sqrt{{{g}^{2}}+{{f}^{2}}-c}$
Now the equation of first circle is given as ${{x}^{2}}+{{y}^{2}}-10x+16y+89-{{r}^{2}}=0$,
So, its centre ${{c}_{1}}=(5,-8)$ and its radius ${{r}_{1}}=\sqrt{{{(5)}^{2}}+{{(-8)}^{2}}-(89-{{r}^{2}})}$
On simplifying we get,
${{r}_{1}}=r$units
Similarly, equation of second circle is ${{x}^{2}}+{{y}^{2}}+6x-14y+42=0$,
Its centre ${{c}_{2}}=(-3,7)$
And its radius ${{r}_{2}}=\sqrt{{{(-3)}^{2}}+{{(7)}^{2}}-42}$
Simplifying we get,
${{r}_{2}}=4$ units
Now we know that distance formula between the points $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$ is $\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}$.
Therefore, the distance between the centres of two circles is given by,
${{c}_{1}}{{c}_{2}}=\sqrt{{{(5+3)}^{2}}+{{(-8-7)}^{2}}}$
On simplifying we get,
${{c}_{1}}{{c}_{2}}=\sqrt{64+225}$
Adding we get,
${{c}_{1}}{{c}_{2}}=\sqrt{289}$
On taking positive square root we get,
${{c}_{1}}{{c}_{2}}=17$ units
Now, if these two circles intersect the sum of their radii must be greater than the distance between their centres. In that case the number of tangents will be $2$.
${{r}_{1}}+{{r}_{2}}>17$
On substituting the value we get,
$r+4>17$
Simplifying we get,
$r>13$
Since, for $r$ equal to $21$, the number of common tangents is $2$.
Again the number of common tangents are $3$ if and only if ,
${{r}_{1}}+{{r}_{2}}=17$
$r+4=17$
Simplifying we get,
$r=13$ units
Therefore, for $r=13$ units the common tangents are $3$.

Therefore, the correct options are (C) and (D).

Note:
There are $3$ cases:
Case 1: If these two circles intersect the sum of their radii must be greater than the distance between their centres. In that case the number of tangents will be $2$.
Case 2: If two circles meet at single point, then the sum of their radii must be equal to the distance between their centres. In that case the number of common tangents are $3$.
Case 3: If two circles neither meet and nor intersect then the sum of their radii are less than the distance between their centres. In this case, the number of common tangents are $4$.