Question

The tangent to the curve $$y=x^2+6$$ at a point P (1, 7) touches the circle $$x^2+y^2+16x+12y+c=0$$ at a point Q. Then the coordinates of Q are: -
(a) (-6, -7)
(b) (-10, -15)
(c) (-9, -13)
(d) (-6, -11)

Answer 1

Hint:

First of all find the slope of the tangent to the curve $$y=x^2+6$$ at a point P (1, 7) by differentiating both sides of the curve with respect to x and substituting the values of coordinates. Now, apply the formula: - $$(y-y_1)=m(x-x_1)$$ to determine the equation of the tangent line. Here, ‘m’ is the slope given as $${\displaystyle \frac{dy}{dx}}$$ at point P (1, 7), $$(x_1,y_1)$$ is the point P (1, 7). Now, solve the equation of the tangent and the circle and substitute the value of discriminant equal to 0, of the obtained quadratic equation. Finally, assume the coordinates of Q as (x, y) and use the formula: - $${\displaystyle \frac{\alpha -x}{a}}={\displaystyle \frac{\beta -y}{b}}=\left({\displaystyle \frac{a\alpha +b\beta +c^{\prime }}{a^2+b^2}}\right)$$ to solve for the values of x and y. Here, $$(\alpha ,\beta )=(-g,-f)$$ = coordinates of the centre of the circle, a, b and c’ are the coefficients of x, y and constant term in the equation of tangent.

Complete step by step answer:

Here, we have been given that the tangent of the curve $$y=x^2+6$$ is touching the given circle at point Q and we have to determine the coordinates of Q.

Now, first let us determine the equation of tangent to the curve $$y=x^2+6$$ by determining the slope on differentiating the curves with respect to x,

$\Rightarrow {\displaystyle \frac{dy}{dx}}=2x+0$

 $\Rightarrow {\displaystyle \frac{dy}{dx}}=2x$

So, at point P (1, 7), we have,

$$\Rightarrow {\displaystyle \frac{dy}{dx}}=2\times 1=2$$

Here, $${\displaystyle \frac{dy}{dx}}$$ denotes the slope of the tangent to the curve. Let us assume it as ‘m’. So, we have,

$$\Rightarrow m=2$$

Using the general equation of a straight line given as: - $$(y-y_1)=m(x-x_1)$$, where $$(x_1,y_1)=(1,7)$$, we get,

$\Rightarrow (y-7)=2(x-1)$

 $\Rightarrow y-7=2x-2$

$$\Rightarrow y=2x+5$$ - (1)

So, the above equation represents the tangent line to the curve $$y=x^2+6$$ at point P (1, 7).

Now, it is said that this line touches the circle $$x^2+y^2+16x+12y+c=0$$ at point Q. Comparing the given equation of the circle with the general equation of the circle given as: - $$x^2+y^2+2gx+2fy+c=0$$, we get,

$$\Rightarrow 2g=16$$ and $$2f=12$$

$$\Rightarrow g=8$$ and $$f=6$$

Since, the centre of the circle is given as: - $$(-g,-f)$$. So, we have (-8, -6) as the centre of the provided circle. Let us draw a diagram of the given situation.

Now, in the above figure we can see that the line touches the circle at only one point so they must have a solution. So, solving the two equations of substituting the value of y from equation (1) in the equation of circle, we get,

$\Rightarrow x^2+{(2x+5)}^2+16x+12(2x+5)+c=0$

$\Rightarrow x^2+4x^2+25+20x+16x+24x+60+c=0$

$\Rightarrow 5x^2+60x+85+c=0$

$\Rightarrow x^2+12x+(17+{\displaystyle \frac{c}{5}})=0$

We know that a quadratic equation, $$A{x}^2+Bx+C=0$$ will have only one solution if its discriminant will be 0. So, we get,

$$\Rightarrow$$ D = Discriminant = $$B^2-4AC=0$$

Here, B = 12, A = 1 and C = $$(17+{\displaystyle \frac{c}{5}})$$

$\Rightarrow {12}^2-4\times 1(17+{\displaystyle \frac{c}{5}})=0$

 $\Rightarrow 144-(68+{\displaystyle \frac{4c}{5}})=0$

 $\Rightarrow 144-68-{\displaystyle \frac{4c}{5}}=0$

 $\Rightarrow {\displaystyle \frac{4c}{5}}=76$

 $\Rightarrow c=95$

Therefore, the equation of the circle is $$x^2+y^2+16x+12y+95=0$$.

Now, we know that radius is perpendicular to the tangent at the point of constant. So, OQ is perpendicular to the tangent line, y = 2x + 5. In the figure, we have assumed the foot of perpendicular as Q (x, y). We know that if a perpendicular is drawn from a point $$(\alpha ,\beta )$$ to the line $$ax+by+c^{\prime }=0$$ then the coordinates of the foot is given as: -

$${\displaystyle \frac{\alpha -x}{a}}={\displaystyle \frac{\beta -y}{b}}=\left({\displaystyle \frac{a\alpha +b\beta +c}{a^2+b^2}}\right)$$

So, converting the equation of tangent in the form $$ax+by+c^{\prime }=0$$, we get,

$$\Rightarrow 2x-y+5=0$$, here a = 2, b = -1, c’ = 5

Since, perpendicular is drawn from point O (-8, -6). So, we have,

$$\Rightarrow (\alpha ,\beta )=(-8,-6)$$

Substituting all the values in the formula, we get,

 $\Rightarrow {\displaystyle \frac{-8-x}{2}}={\displaystyle \frac{-6-y}{-1}}=\left({\displaystyle \frac{2\times (-8)+(-1)\times (-6)+5}{2^2+{(-1)}^2}}\right)$

 $\Rightarrow {\displaystyle \frac{-8-x}{2}}=6+y=\left({\displaystyle \frac{-16+6+5}{4+1}}\right)$

 $\Rightarrow {\displaystyle \frac{-8-x}{2}}=6+y={\displaystyle \frac{-5}{5}}$

 $\Rightarrow {\displaystyle \frac{-8-x}{2}}=6+y=-1$

\end{align}\]

Considering, $${\displaystyle \frac{-8-x}{2}}=-1$$ we get,

$\Rightarrow -8-x=-2$

 $\Rightarrow x=-6$

Considering $$6+y=-1$$, we get,

$$\Rightarrow y=-7$$

Here, the coordinates of point Q is (-6, -7). Therefore, option (a) is the correct answer.

Note:

 One may note that we must determine the value of constant ‘c’ in the equation of a circle to reach our answer. You must remember the condition of a quadratic equation having only one solution and the formulas for determining the coordinates of the foot of perpendicular. Do not forget to draw the diagram as it will help you to understand the situation clearly. You can also apply the formula: - $$r=\sqrt{g^2+f^2-c}=\left|{\displaystyle \frac{a\alpha +\beta b+c}{\sqrt{a^2+b^2}}}\right|$$ to determine the value of c.