Question

The tangent to the parabola $y=x^2+6$ at a point P (1, 7) touches the circle $x^2+y^2+16x+12y+c=0$ at a point Q. Then the co-ordinates of Q are
(a) (– 6, – 9)
(b) (– 13, – 9)
(c) (– 6, – 7)
(d) (13, 7)

Answer 1

Hint:In this question, firstly, find the equation of the tangent by differentiating the equation of parabola with respect to $x$ and using the point-slope formula $y-y_1=m(x-x_1)$ . The next step is to find the center of the circle from the given equation of the circle by comparing it to the general form of the equation of the circle $x^2+y^2+2gx+2fy+c=0$ which gives the center $(-g,-f)$ . Now, draw the diagram with the obtained data and find the radius or the distance between the line and the center by using the formula of distance from the point to the line, $d={\displaystyle \frac{|a{x}_1+b{y}_1+c|}{\sqrt{a^2+b^2}}}$ . Finally, use the trial and error method and use the formula of distance between two points, $d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$ for every option mentioned in the question. If the radius of the circle is equal to the distance between the two points ( center of the circle and the intersection point of circle and tangent) then that option is your correct answer.

Complete step by step answer:
$\Rightarrow$ We have been given the equation of the parabola, let us first find the equation of the tangent at P (1, 7).
We know, the point slope form of the equation, $y-y_1=m(x-x_1)$
To find the slope of the equation, we will differentiate the equation of the parabola with respect to $x$ , we get
$\begin{array}{rl}& {\displaystyle \frac{d}{dx}}\left(y\right)={\displaystyle \frac{d}{dx}}(x^2+6)\\ & {\displaystyle \frac{dy}{dx}}=2x\end{array}$
We know that, $m={\displaystyle \frac{dy}{dx}}$ , at point P (1, 7), we get
$\begin{array}{rl}& m=2\left(1\right)\\ & m=2\end{array}$
Now, let us use the point-slope form of the equation which has $m=2$ and P (1, 7) to find the equation of the tangent.
$\begin{array}{rl}& y-7=2(x-1)\\ & y-7=2x-2\\ & y=2x-2+7\end{array}$
$\Rightarrow$ $y=2x+5$ …………….. (1)
For the next step, we have the equation of the circle,
 $x^2+y^2+16x+12y+c=0$
The standard form of the circle is $x^2+y^2+2gx+2fy+c=0$ where the center of the circle is $(-g,-f)$
Therefore, the center of the given circle,$x^2+y^2+16x+12y+c=0$
$\begin{array}{rl}& 2g=16\\ & g=8\end{array}$
$\begin{array}{rl}& 2f=12\\ & f=6\end{array}$
$\Rightarrow$ Now, let center of the circle be O $(-g,-f)$ = O (– 8, – 6)
Let us draw the diagram from the obtained data and the given data for the parabola, circle and the tangent.

$\Rightarrow$ So, the tangent that touches the parabola at P (1, 7) also touches the circle with a center O (– 8, – 6) at the point Q (x, y).
If we look closer to the circle and the tangent,

$\Rightarrow$ We can see that the tangent is normal to the radius of the circle, therefore, let us find the distance from the center of the circle to the tangent which is also the radius.
$\Rightarrow$ We know, the distance from a point to the line is given by,
$d={\displaystyle \frac{|a{x}_1+b{y}_1+c|}{\sqrt{a^2+b^2}}}$
where, $d$ is the distance from the point $(x_1,y_1)$ to the line $ax+by+c=0$
$\Rightarrow$ From the above diagram we have, $(x_1,y_1)=\text{O}(-8,-6)$ and the tangent $y=2x+5$ is the line, the distance from the point to line, $d=r$ .
$\begin{array}{rl}& r={\displaystyle \frac{|\left(2\right)(-8)+(-1)(-6)+5|}{\sqrt{2^2+{(-1)}^2}}}\\ & ={\displaystyle \frac{|-16+6+5|}{\sqrt{4+1}}}\\ & ={\displaystyle \frac{|-16+11|}{\sqrt{5}}}\\ & ={\displaystyle \frac{|-5|}{\sqrt{5}}}\\ & ={\displaystyle \frac{5}{\sqrt{5}}}\\ & =\sqrt{5}\end{array}$
$\Rightarrow$ Therefore, the radius of the circle is $\sqrt{5}$ .
Now, we know the distance between the two points, Q and O. So, let us use the distance formula between two points, and with the help of the trial and error method check which of the options is correct.
We know,
$\Rightarrow$ Distance between two points, $d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$
(a) (– 6, – 9):
$\Rightarrow$ We have two points Q (– 6, – 9) and O (– 8, – 6)
$\begin{array}{rl}& d=\sqrt{{(-8-(-6))}^2+{(-6-(-9))}^2}\\ & =\sqrt{{(-8+6)}^2+{(-6+9)}^2}\\ & =\sqrt{{(-2)}^2+{\left(3\right)}^2}\\ & =\sqrt{4+9}\\ & =\sqrt{13}\end{array}$
$\Rightarrow$ Since, $d=\sqrt{13}$ is not equal to $r=\sqrt{5}$ , the first option is not the correct answer.
$\Rightarrow$ Similarly, if we check for the second option, we will not get the distance and the radius equal.
(c) (– 6, – 7):
$\Rightarrow$ We have two points O (– 8, – 6) and Q (– 6, – 7)
$\begin{array}{rl}& d=\sqrt{{(-6-(-8))}^2+{(-7-(-6))}^2}\\ & =\sqrt{{(-6+8)}^2+{(-7+6)}^2}\\ & =\sqrt{{\left(2\right)}^2+{(-1)}^2}\\ & =\sqrt{4+1}\\ & =\sqrt{5}\end{array}$
$\Rightarrow$ Here, from the distance formula we got the distance between the two point and the radius of the circle equal.
$\Rightarrow$ Hence, the coordinates of Q are (– 6, – 7).
Note:
Here, there is another way to solve this question, simply use the trial and error method by substituting each option in the equation of tangent and if the total value equals to 0, then that relative point taken to get the value 0 is the correct answer.