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Question: If the tangent at \((1,7)\) to the curve \({x^2} = y - 6\) touches the circle \({x^2} + {y^2} + 16x ...

If the tangent at (1,7)(1,7) to the curve x2=y−6{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
(1)85(1)85
(2)95(2)95
(3)195(3)195
(4)185(4)185

Explanation

Solution

First, we have to define what is slope; the slope is calculated by finding the ratio of the vertical changes in the given x and y-axis and also finding the horizontal changes between any two distinct points on the line.
The vertical changes between the given two points are called the rise, and horizontal changes are known as the run.
Formula used: Tangent of the slope m=∣dydx∣(x,y)m = {\left| {\dfrac{{dy}}{{dx}}} \right|_{(x,y)}}

Complete step-by-step solution:
Let from the given equation the curve equation is x2=y−6{x^2} = y - 6.
Now convert the equation into single degree y on the left side, so that we are able to differentiate with the value y to simplify the solution.
Thus, we get x2=y−6⇒y=x2+6{x^2} = y - 6 \Rightarrow y = {x^2} + 6.
Differentiate the value y with base x, thus we get dydx=2x\dfrac{{dy}}{{dx}} = 2x.
Now we are going to apply the formula of the solution to the diffraction of the function thus we get,
m=∣dydx∣(1,7)⇒2xm = {\left| {\dfrac{{dy}}{{dx}}} \right|_{(1,7)}} \Rightarrow 2x (since on the right side of the equation there are only x terms, we are applying the x terms as x is one)
Thus, we get, m⇒2(1)=2m \Rightarrow 2(1) = 2
Since the equation of the tangent points is (1,7)(1,7) apply in the given equation we get y−7=2(x−1)y - 7 = 2(x - 1)
Solving the equation, we get, y−7=2(x−1)⇒2x−y+5=0y - 7 = 2(x - 1) \Rightarrow 2x - y + 5 = 0
From the given circle equation as x2+y2+16x+12y+c=0{x^2} + {y^2} + 16x + 12y + c = 0 convert this equation according to the plane touches thus we get, (x+8)2+(y+6)2+c−64−36=0{(x + 8)^2} + {(y + 6)^2} + c - 64 - 36 = 0
If we simplify the terms and cancel the common values, we get the given,
Thus, we get, (x+8)2+(y+6)2=100−c{(x + 8)^2} + {(y + 6)^2} = 100 - c
Since from this equation, we get the distance as (−8,−6)( - 8, - 6) from the equation 2x−y+5=02x - y + 5 = 0
Hence the distance is d=∣2(−8)−(−6)4+1∣d = \left| {\dfrac{{2( - 8) - ( - 6)}}{{\sqrt {4 + 1} }}} \right| after simplifying the terms we get, d=∣5∣d = \left| {\sqrt 5 } \right|
Now equivalent this to the radius we get, Radius 100−c=5\sqrt {100 - c} = \sqrt 5 where c is the constant
Thus, we get c=95c = 95 (is the value of the c is ninety-five then both equations are equal)
Hence the value of c is 9595. And this option(2)95(2)95 is correct.
There is no possibility of getting other options because if the value of c is not 9595 then the radius value will be not equivalent 100−c=5\sqrt {100 - c} = \sqrt 5 .

Note: Tangent is the line that the plane curves at the given points in the straight line which gets touched into the curve plane.
Where c is the constant the value of the c does not affect the square root.
m=∣dydx∣(x,y)m = {\left| {\dfrac{{dy}}{{dx}}} \right|_{(x,y)}} is the general formula for slope where x and y are the axes.