# If the tangent at \[\left( 1,7 \right)\] to the curve \[{{x}^{2}}=y-6\] touches the circle \[{{x}^{2}}+{{y}^{2}}+16x+12y+c=0\] then the value of \[c\] is

(a) \[85\]

(b) \[95\]

(c) \[195\]

(d) \[185\]

Last updated date: 17th Mar 2023

•

Total views: 304.2k

•

Views today: 7.83k

Answer

Verified

304.2k+ views

Hint: Write the equation of tangent at point \[\left( 1,7 \right)\] to the curve \[{{x}^{2}}=y-6\] using point slope form by writing the slope in terms of \[\dfrac{dy}{dx}\]. Assume a point on the circle at which the tangent touches the circle and write the slope of the line joining the centre of the circle to this point. This line is perpendicular to the tangent and thus, the product of their slopes is \[-1\]. Use this fact to find the point at which tangent touches the circle. Substitute the value of this point in the equation of the circle to find the value of \[c\].

Complete step-by-step answer:

We have a curve of the form \[{{x}^{2}}=y-6\] to which tangent is drawn at point \[\left( 1,7 \right)\]. This tangent also touches the circle \[{{x}^{2}}+{{y}^{2}}+16x+12y+c=0\]. We have to find the value of parameter \[c\].

We will first find the equation of tangent to the curve \[{{x}^{2}}=y-6\] at point \[\left( 1,7 \right)\]. We can write the slope of the curve as \[\dfrac{dy}{dx}\]. Thus, differentiating the equation \[{{x}^{2}}=y-6\] on both sides, we have \[2xdx=dy\] as the derivative of any function of the form \[y=a{{x}^{n}}+b\] is \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].

Thus, we have \[\dfrac{dy}{dx}=2x\]. Substituting the point \[\left( 1,7 \right)\] in the equation, we have \[\dfrac{dy}{dx}=2\] as the slope of the tangent at point \[\left( 1,7 \right)\].

We will write the equation of tangent. We know that any line passing through point \[\left( {{x}_{1}},{{y}_{1}} \right)\] having slope \[m\] is \[y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)\].

Substituting \[{{x}_{1}}=1,{{y}_{1}}=7,m=2\] in the above equation, we have \[y-7=2\left( x-1 \right)\Rightarrow y=2x+5\] as the equation of line passing through \[\left( 1,7 \right)\] having slope \[2\].

The tangent with equation \[y=2x+5\] is also tangent to the circle \[{{x}^{2}}+{{y}^{2}}+16x+12y+c=0\].

Let’s assume that tangent touches the circle at point \[\left( u,v \right)\].

Thus, the line joining centre of the circle to point \[\left( u,v \right)\] is perpendicular to the equation of tangent \[y=2x+5\].

We know that a circle with equation \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\] has centre at point \[\left( -g,-f \right)\]. Thus, the centre of the circle \[{{x}^{2}}+{{y}^{2}}+16x+12y+c=0\] is \[\left( -8,-6 \right)\].

We will find the slope of the line joining the centre of the circle \[\left( -8,-6 \right)\] to the point \[\left( u,v \right)\].

We know that the slope of line joining two points \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] is \[\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\].

Substituting \[{{x}_{1}}=-8,{{y}_{1}}=-6,{{x}_{2}}=u,{{y}_{2}}=v\] in the above equation, we have \[\dfrac{v+6}{u+8}\] as the slope of line joining the centre of circle \[\left( -8,-6 \right)\] to the point \[\left( u,v \right)\]. We know this line is perpendicular to the equation of tangent.

We know that the product of slopes of two perpendicular lines is \[-1\].

As the slope of the tangent is \[2\], we have \[\dfrac{v+6}{u+8}\left( 2 \right)=-1\].

Simplifying the above equation, we have \[2v+12=-u-8\].

\[\Rightarrow 2v+u+20=0.....\left( 1 \right)\]

We know that the point \[\left( u,v \right)\] passes through the equation of tangent \[y=2x+5\]. Thus, we have \[v=2u+5.....\left( 2 \right)\].

Substituting equation \[\left( 2 \right)\] in equation \[\left( 1 \right)\], we have \[4u+10+u+20=0\].

\[\begin{align}

& \Rightarrow 5u+30=0 \\

& \Rightarrow u=-6 \\

\end{align}\]

Substituting the value \[u=-6\] in equation \[\left( 1 \right)\], we have \[2v-6+20=0\].

\[\begin{align}

& \Rightarrow 2v+14=0 \\

& \Rightarrow v=-7 \\

\end{align}\]

Thus, we have \[\left( u,v \right)=\left( -6,-7 \right)\] as the coordinates of the point at which the tangent touches the circle. Substituting this point \[\left( u,v \right)=\left( -6,-7 \right)\] in the equation of circle \[{{x}^{2}}+{{y}^{2}}+16x+12y+c=0\], we have \[{{\left( -6 \right)}^{2}}+{{\left( -7 \right)}^{2}}+16\left( -6 \right)+12\left( -7 \right)+c=0\].

Simplifying the equation, we have \[-95+c=0\].

\[\Rightarrow c=95\]

Hence, the value of \[c\] for which the tangent to the parabola is tangent to the circle as well is \[c=95\] which is option (b).

Note: We can also solve this question by finding the foot of perpendicular drawn from the centre of the circle to the equation of tangent and then substituting the value of foot of perpendicular in the equation of circle to get the exact equation of circle.

Complete step-by-step answer:

We have a curve of the form \[{{x}^{2}}=y-6\] to which tangent is drawn at point \[\left( 1,7 \right)\]. This tangent also touches the circle \[{{x}^{2}}+{{y}^{2}}+16x+12y+c=0\]. We have to find the value of parameter \[c\].

We will first find the equation of tangent to the curve \[{{x}^{2}}=y-6\] at point \[\left( 1,7 \right)\]. We can write the slope of the curve as \[\dfrac{dy}{dx}\]. Thus, differentiating the equation \[{{x}^{2}}=y-6\] on both sides, we have \[2xdx=dy\] as the derivative of any function of the form \[y=a{{x}^{n}}+b\] is \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].

Thus, we have \[\dfrac{dy}{dx}=2x\]. Substituting the point \[\left( 1,7 \right)\] in the equation, we have \[\dfrac{dy}{dx}=2\] as the slope of the tangent at point \[\left( 1,7 \right)\].

We will write the equation of tangent. We know that any line passing through point \[\left( {{x}_{1}},{{y}_{1}} \right)\] having slope \[m\] is \[y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)\].

Substituting \[{{x}_{1}}=1,{{y}_{1}}=7,m=2\] in the above equation, we have \[y-7=2\left( x-1 \right)\Rightarrow y=2x+5\] as the equation of line passing through \[\left( 1,7 \right)\] having slope \[2\].

The tangent with equation \[y=2x+5\] is also tangent to the circle \[{{x}^{2}}+{{y}^{2}}+16x+12y+c=0\].

Let’s assume that tangent touches the circle at point \[\left( u,v \right)\].

Thus, the line joining centre of the circle to point \[\left( u,v \right)\] is perpendicular to the equation of tangent \[y=2x+5\].

We know that a circle with equation \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\] has centre at point \[\left( -g,-f \right)\]. Thus, the centre of the circle \[{{x}^{2}}+{{y}^{2}}+16x+12y+c=0\] is \[\left( -8,-6 \right)\].

We will find the slope of the line joining the centre of the circle \[\left( -8,-6 \right)\] to the point \[\left( u,v \right)\].

We know that the slope of line joining two points \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] is \[\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\].

Substituting \[{{x}_{1}}=-8,{{y}_{1}}=-6,{{x}_{2}}=u,{{y}_{2}}=v\] in the above equation, we have \[\dfrac{v+6}{u+8}\] as the slope of line joining the centre of circle \[\left( -8,-6 \right)\] to the point \[\left( u,v \right)\]. We know this line is perpendicular to the equation of tangent.

We know that the product of slopes of two perpendicular lines is \[-1\].

As the slope of the tangent is \[2\], we have \[\dfrac{v+6}{u+8}\left( 2 \right)=-1\].

Simplifying the above equation, we have \[2v+12=-u-8\].

\[\Rightarrow 2v+u+20=0.....\left( 1 \right)\]

We know that the point \[\left( u,v \right)\] passes through the equation of tangent \[y=2x+5\]. Thus, we have \[v=2u+5.....\left( 2 \right)\].

Substituting equation \[\left( 2 \right)\] in equation \[\left( 1 \right)\], we have \[4u+10+u+20=0\].

\[\begin{align}

& \Rightarrow 5u+30=0 \\

& \Rightarrow u=-6 \\

\end{align}\]

Substituting the value \[u=-6\] in equation \[\left( 1 \right)\], we have \[2v-6+20=0\].

\[\begin{align}

& \Rightarrow 2v+14=0 \\

& \Rightarrow v=-7 \\

\end{align}\]

Thus, we have \[\left( u,v \right)=\left( -6,-7 \right)\] as the coordinates of the point at which the tangent touches the circle. Substituting this point \[\left( u,v \right)=\left( -6,-7 \right)\] in the equation of circle \[{{x}^{2}}+{{y}^{2}}+16x+12y+c=0\], we have \[{{\left( -6 \right)}^{2}}+{{\left( -7 \right)}^{2}}+16\left( -6 \right)+12\left( -7 \right)+c=0\].

Simplifying the equation, we have \[-95+c=0\].

\[\Rightarrow c=95\]

Hence, the value of \[c\] for which the tangent to the parabola is tangent to the circle as well is \[c=95\] which is option (b).

Note: We can also solve this question by finding the foot of perpendicular drawn from the centre of the circle to the equation of tangent and then substituting the value of foot of perpendicular in the equation of circle to get the exact equation of circle.

Recently Updated Pages

If ab and c are unit vectors then left ab2 right+bc2+ca2 class 12 maths JEE_Main

A rod AB of length 4 units moves horizontally when class 11 maths JEE_Main

Evaluate the value of intlimits0pi cos 3xdx A 0 B 1 class 12 maths JEE_Main

Which of the following is correct 1 nleft S cup T right class 10 maths JEE_Main

What is the area of the triangle with vertices Aleft class 11 maths JEE_Main

The coordinates of the points A and B are a0 and a0 class 11 maths JEE_Main

Trending doubts

Write an application to the principal requesting five class 10 english CBSE

Tropic of Cancer passes through how many states? Name them.

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE

What is per capita income

Change the following sentences into negative and interrogative class 10 english CBSE

A Short Paragraph on our Country India