Question

If $16{{l}^{2}}+9{{m}^{2}}=24lm+6l+8m+1$ and if S is the equation of the circle having $lx+my+1=0$ as a tangent , then the equation of the director circle of S is
[a] ${{x}^{2}}+{{y}^{2}}+6x+8y=25$
[b] ${{x}^{2}}+{{y}^{2}}-6x+8y=25$
[c] ${{x}^{2}}+{{y}^{2}}+6x+8y=25$
[d] ${{x}^{2}}+{{y}^{2}}+6x-8y=25$

Answer 1

Hint: Assume that the centre of the circle is (a,b) and let the radius of the circle is r. Use the fact that the distance of the line lx+my +1 = 0 from the centre of the circle is equal r and hence prove that $\dfrac{\left| al+bm+1 \right|}{\sqrt{{{l}^{2}}+{{m}^{2}}}}=r$. Square both sides of the equation and compare the equation with $16{{l}^{2}}+9{{m}^{2}}=24lm+6l+8m+1$ and hence find the value of a, b and r. Hence find the equation of the circle. Use the fact that the equation of the director circle of the circle ${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}$ is ${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}=2{{r}^{2}}$. Hence find the equation of the director circle and hence determine which of the options is correct.

Complete step by step answer:

Let the centre of the circle is $\left( a,b \right)$ and let the radius of the circle is $r$.
Since $lx+my+1=0$ is tangent to the circle, the distance of the center (a,b) from the line is r.
We know that the distance of point $\left( {{x}_{1}},{{y}_{1}} \right)$ from the line $Ax+By+C=0$ is given by $\dfrac{\left| A{{x}_{1}}+B{{y}_{1}}+C \right|}{\sqrt{{{A}^{2}}+{{B}^{2}}}}$
Hence, we have
$\dfrac{\left| al+bm+1 \right|}{\sqrt{{{l}^{2}}+{{m}^{2}}}}=r$
Multiplying both sides by $\sqrt{{{l}^{2}}+{{m}^{2}}}$, we get
$\left| al+bm+1 \right|=r\sqrt{{{l}^{2}}+{{m}^{2}}}$
Squaring both sides, we get
${{\left( al+bm+1 \right)}^{2}}={{r}^{2}}\left( {{l}^{2}}+{{m}^{2}} \right)$
We know that ${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac$
Hence, we have
${{\left( al+bm+1 \right)}^{2}}={{\left( al \right)}^{2}}+{{\left( bm \right)}^{2}}+{{1}^{2}}+2\left( al \right)\left( bm \right)+2\left( bm \right)\left( 1 \right)+2\left( 1 \right)\left( al \right)$
Hence, we have
\[{{\left( al \right)}^{2}}+{{\left( bm \right)}^{2}}+{{1}^{2}}+2\left( al \right)\left( bm \right)+2\left( bm \right)\left( 1 \right)+2\left( 1 \right)\left( al \right)={{r}^{2}}{{l}^{2}}+{{r}^{2}}{{m}^{2}}\]
Subtracting ${{a}^{2}}{{l}^{2}}+{{b}^{2}}{{m}^{2}}$ from both sides, we get
${{l}^{2}}\left( {{r}^{2}}-{{a}^{2}} \right)+{{m}^{2}}\left( {{r}^{2}}-{{b}^{2}} \right)=2ablm+2al+2bm+1$
Comparing with $16{{l}^{2}}+9{{m}^{2}}=24lm+6l+8m+1$, we get
On comparing coefficients of l, we get
$2a=6\Rightarrow a=3$
On comparing coefficients of m, we get
$2b=8\Rightarrow b=4$
Comparing coefficients of ${{l}^{2}}$, we get
${{r}^{2}}-{{a}^{2}}=16\Rightarrow {{r}^{2}}-9=16\Rightarrow {{r}^{2}}=25\Rightarrow r=5$
Hence, the centre of the circle is $\left( 3,4 \right)$ and the radius of the circle is $5$
Hence, the equation of the circle is ${{\left( x-3 \right)}^{2}}+{{\left( y-4 \right)}^{2}}={{5}^{2}}$
We know that the equation of the director circle of the circle ${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}$ is ${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}=2{{r}^{2}}$.
Hence the equation of the director circle is
${{\left( x-3 \right)}^{2}}+{{\left( y-4 \right)}^{2}}=2\times {{5}^{2}}=50$
We know that ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$
Hence, we have
$\begin{align}
  & {{x}^{2}}-6x+9+{{y}^{2}}-8y+16=50 \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}-6x-8y=25 \\
\end{align}$
Hence option [c] is correct.

Note:
[1] Verification: We can verify the correctness of our solution by checking that the condition of tangency of line $lx+my+1=0$ to the circle ${{\left( x-3 \right)}^{2}}+{{\left( y-4 \right)}^{2}}=25$ is the equation $16{{l}^{2}}+9{{m}^{2}}=24lm+6l+8m+1$
The distance of (3,4) from $lx+my+1=0$ is
$\dfrac{\left| 3l+4m+1 \right|}{\sqrt{{{l}^{2}}+{{m}^{2}}}}$
The line $lx+my+1=0$ will be tangent to the circle if this distance is equal to the radius, i.e. 5
Hence, we have
$\begin{align}
  & \dfrac{\left| 3l+4m+1 \right|}{\sqrt{{{l}^{2}}+{{m}^{2}}}}=5 \\
 & \Rightarrow \left| 3l+4m+1 \right|=5\sqrt{{{l}^{2}}+{{m}^{2}}} \\
\end{align}$
Squaring both sides and using the identity ${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac$, we get
$9{{l}^{2}}+16{{m}^{2}}+1+6l+8m+24lm=25{{l}^{2}}+25{{m}^{2}}$
Subtracting $9{{l}^{2}}+16{{m}^{2}}$ on both sides, we get
$16{{l}^{2}}+9{{m}^{2}}=24lm+6l+8m+1$
Hence our solution is verified to be correct.