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Question: If the tangent at \[\left( 1,7 \right)~\] to the curve \[{{x}^{2}}=y-6~\] touches the circle \[{{x}^...

If the tangent at (1,7) \left( 1,7 \right)~ to the curve x2=y6 {{x}^{2}}=y-6~ touches the circle x2+y2+16x+12y+c=0 {{x}^{2}}+{{y}^{2}}+16x+12y+c=0~ then the value of cc is
A) 85
B) 95
C) 195
D) 185

Explanation

Solution

We will find the slope of the tangent to the given curve and then we will find the equation of the tangent using the slope of the tangent and the given point on the tangent. Then we will use this equation of tangent to find the distance of the center of the circle to the tangent and we will equate the obtained distance with the radius of the circle.

Complete step by step solution:
The given equation of the curve is x2=y6 {{x}^{2}}=y-6~.
We can write the equation of curve as
y=x2+6 y={{x}^{2}}+6~ …….. (1)\left( 1 \right)
Now, we will find the slope of the tangent to the curve at (1,7) \left( 1,7 \right)~ by differentiating the equation of the curve with respect to xx.
slope=(dydx)(1,7) \Rightarrow slope={{\left( \dfrac{dy}{dx} \right)}_{\left( 1,7 \right)~}}
Now, we will substitute the value of from equation 1, we get
slope=(d(x2+6) dx)(1,7) \Rightarrow slope={{\left( \dfrac{d\left( {{x}^{2}}+6 \right)~}{dx} \right)}_{\left( 1,7 \right)~}}
On differentiating the terms, we get
slope=(2x)(1,7) \Rightarrow slope={{\left( 2x \right)}_{\left( 1,7 \right)~}}
Substituting the value of x coordinate of point, we get
slope=2×1=2\Rightarrow slope=2\times 1=2
Now, we will find the equation of the tangent to the curve.
y7=2(x1)\Rightarrow y-7=2\left( x-1 \right)
Simplifying the terms further, we get
y7=2x2\Rightarrow y-7=2x-2
Adding 7 on both sides of the equation, we get
y7+7=2x2+7 y=2x+5 \begin{aligned} & \Rightarrow y-7+7=2x-2+7 \\\ & \Rightarrow y=2x+5 \\\ \end{aligned}.............. (2)\left( 2 \right)
Given equation of the circle is
x2+y2+16x+12y+c=0 \Rightarrow {{x}^{2}}+{{y}^{2}}+16x+12y+c=0~
We can also write the equation of circle as
(x+8)2+(y+6)2+c6436=0 \Rightarrow {{\left( x+8 \right)}^{2}}+{{\left( y+6 \right)}^{2}}+c-64-36=0~
On further simplification, we get
(x+8)2+(y+6)2=100 c\Rightarrow {{\left( x+8 \right)}^{2}}+{{\left( y+6 \right)}^{2}}=100~-c
We can see from the equation of the circle that the point of the center of the circle is (8,6)\left( -8,-6 \right).
If the tangent touches the circle, then the distance of the tangent from the center is equal to the radius of the circle.
Also, the distance of point (8,6)\left( -8,-6 \right) from y=2x+5y=2x+5 is equal to the radius of the circle.
2(8)(6)4+1=100c\Rightarrow \left| \dfrac{2\left( -8 \right)-\left( -6 \right)}{\sqrt{4+1}} \right|=\left| \sqrt{100-c} \right|
On simplifying the terms, we get
5=100c\Rightarrow \left| \sqrt{5} \right|=\left| \sqrt{100-c} \right|
Squaring both sides, we get
52=100c2 5=100c \begin{aligned} & \Rightarrow {{\left| \sqrt{5} \right|}^{2}}={{\left| \sqrt{100-c} \right|}^{2}} \\\ & \Rightarrow 5=100-c \\\ \end{aligned}
On further simplification, we get
c=95\Rightarrow c=95

Hence, the correct option is option B.

Note:
Definition and some important properties of tangent to a curve are as follows:-
A tangent is a line that meets a curve at a point but does not cross/intercept with it so that the gradient of the point where the tangent touches is the same with that of the line. A tangent can only be a line. Only curves can have tangents and also the curve has a constantly changing gradient for each point.