Question

Tangent to the parabola $y={{x}^{2}}+6$ at $\left( 1,7 \right)$ touches the circle ${{x}^{2}}+{{y}^{2}}+16x+12y+c=0$ at
(a) \[\left( -6,-9 \right)\]
(b) \[\left( -13,-9 \right)\]
(c) $\left( -6,-7 \right)$
(d) $\left( 13,7 \right)$

Answer 1

Hint: First of all, we have to convert the given equation of circle into general form ${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}$. We will obtain the radius in terms of c. Then we have to substitute the given point $\left( 1,7 \right)$ in the equation of the tangent to the parabola \[x{{x}_{1}}=2a\left( y+{{y}_{1}} \right)\] and obtain the slope and intercept of the line hence obtained. Then, we know that the equation of the tangent to the circle is given by \[{{c}^{2}}={{r}^{2}}\left( 1+{{m}^{2}} \right)\], so we can substitute the slope, the intercept and the radius. On solving further, we will obtain the value of c. Hence, we will get the equation of the circle.

Complete step-by-step solution -

The equation of the circle given in the question is ${{x}^{2}}+{{y}^{2}}+16x+12y+c=0$.
We have to convert it to the general equation of the circle given by ${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}$ having the center at \[\left( a,b \right)\] and radius as \[r\].
We can use the method of completing the squares. Therefore, we have to add and subtract \[{{8}^{2}}\] and \[{{6}^{2}}\] as below,
$\begin{align}
  & {{x}^{2}}+{{y}^{2}}+16x+12y+c+{{8}^{2}}-{{8}^{2}}+{{6}^{2}}-{{6}^{2}}=0 \\
 & \left( {{x}^{2}}+16x+{{8}^{2}} \right)+\left( {{y}^{2}}+12y+{{6}^{2}} \right)+c-{{8}^{2}}-{{6}^{2}}=0 \\
 & \left( {{x}^{2}}+16x+{{8}^{2}} \right)+\left( {{y}^{2}}+12y+{{6}^{2}} \right)+c-64-36=0 \\
 & \left( {{x}^{2}}+16x+{{8}^{2}} \right)+\left( {{y}^{2}}+12y+{{6}^{2}} \right)=100-c \\
\end{align}$
Since we know that \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\], we get
\[{{\left( x+8 \right)}^{2}}+{{\left( y+6 \right)}^{2}}=100-c\]
Converting the RHS to a square term, we get
\[{{\left( x+8 \right)}^{2}}+{{\left( y+6 \right)}^{2}}={{\left( \sqrt{100-c} \right)}^{2}}\]
Now, we can compare it with the general equation of the circle. So, we will get the center of the circle as $\left( -8,-6 \right)$ and radius as \[\sqrt{100-c}\].
Now, let us consider the given equation of the parabola $y={{x}^{2}}+6$. We have to convert it into the general equation \[{{x}^{2}}=4ay\]. So, we have the equation as ${{x}^{2}}=\left( y-6 \right)$.
From this, we get that it is an upward parabola with vertex \[\left( 0,6 \right)\] and the term $a=\dfrac{1}{4}$.
We know that the equation of the tangent to the parabola \[{{x}^{2}}=4ay\] at a point \[\left( {{x}_{1}},{{y}_{1}} \right)\] is given by
\[x{{x}_{1}}=2a\left( y+{{y}_{1}} \right)\]
Therefore, we can get the equation of the tangent to the parabola ${{x}^{2}}=\left( y-6 \right)$ at $\left( 1,7 \right)$ by substituting \[{{x}_{1}}=1,{{y}_{1}}=7,a=\dfrac{1}{4}\] as,
\[\begin{align}
  & x\left( 1 \right)=2\left( \dfrac{1}{4} \right)\left( y+7 \right)-6 \\
 & x=\dfrac{1}{2}\left( y+7 \right)-6 \\
 & 2x=\left( y+7 \right)-12 \\
 & 2x=y-5 \\
\end{align}\]
Now, the equation of the tangent obtained above is the equation of a line, so we can compare it with the general equation of a line \[y=mx+c\].
On comparing it, we get
$m=2,c=5$
Since we know that the tangent to the parabola touches the circle, it will also be a tangent to the circle.
The equation of tangent to a circle is given by \[{{c}^{2}}={{r}^{2}}\left( 1+{{m}^{2}} \right)\]. On substituting the values we get,
\[\begin{align}
  & {{5}^{2}}={{\left( \sqrt{100-c} \right)}^{2}}\left( 1+{{2}^{2}} \right) \\
 & 25=\left( 100-c \right)\left( 1+4 \right) \\
 & 25=5\left( 100-c \right) \\
 & 5=100-c \\
 & c=100-5 \\
 & c=95 \\
\end{align}\]
Since we have got the value of \[c=95\], the radius of the circle becomes $r=\sqrt{100-c}\Rightarrow \sqrt{100-95}\Rightarrow \sqrt{5}$
Now, we have the equation of the circle as \[{{\left( x+8 \right)}^{2}}+{{\left( y+6 \right)}^{2}}={{\left( \sqrt{5} \right)}^{2}}\Rightarrow {{\left( x+8 \right)}^{2}}+{{\left( y+6 \right)}^{2}}=5\].
Now we can easily calculate the point of contact by substituting the options given in the question in the above equation of the circle.
Substituting \[\left( -6,-9 \right)\], we get
\[\begin{align}
  & {{\left( -6+8 \right)}^{2}}+{{\left( -9+6 \right)}^{2}}=5 \\
 & {{\left( 2 \right)}^{2}}+{{\left( -3 \right)}^{2}}=5 \\
 & 4+9\ne 5 \\
\end{align}\]
Since it is not satisfying, checking the option \[\left( -13,-9 \right)\], we get
\[\begin{align}
  & {{\left( -13+8 \right)}^{2}}+{{\left( -9+6 \right)}^{2}}=5 \\
 & {{\left( -5 \right)}^{2}}+{{\left( -3 \right)}^{2}}=5 \\
 & 25+9\ne 5 \\
\end{align}\]
Since it is not satisfying, checking the option $\left( -6,-7 \right)$, we get
\[\begin{align}
  & {{\left( -6+8 \right)}^{2}}+{{\left( -7+6 \right)}^{2}}=5 \\
 & {{\left( -2 \right)}^{2}}+{{\left( -1 \right)}^{2}}=5 \\
 & 4+1=5 \\
 & 5=5 \\
\end{align}\]
The point $\left( -6,-7 \right)$ satisfies the equation and it is the correct answer.
Checking the option $\left( 13,7 \right)$ to be sure, we get
\[\begin{align}
  & {{\left( 13+8 \right)}^{2}}+{{\left( 7+6 \right)}^{2}}=5 \\
 & {{\left( 21 \right)}^{2}}+{{\left( 13 \right)}^{2}}=5 \\
 & 441+169\ne 5 \\
\end{align}\]
Therefore, we have obtained the correct answer as $\left( -6,-7 \right)$, which is option (c).

Note: There is another approach to solve this question. We know that the tangent is common to both the circle and the parabola, so we can substitute $y={{x}^{2}}+6$ in the equation of the circle ${{x}^{2}}+{{y}^{2}}+16x+12y+c=0$. The value of \[c\] can be found by using the fact that the distance from the centre of the circle $\left( -8,-6 \right)$ to the tangent to parabola \[2x=y-5\] represents the radius of the circle. The roots of the so formed quadratic equation would lead to the final answer.