Question

If the tangent to the curve \(y^{2}=4x\) at \((4,4)\) touches the circle \(x^{2}+y^{2}-c=0\) then the value of \(5c\) is

• A. \(64\)
• B. \(16\)
• C. \(4\)
• D. \(8\)

Hint:

Find the tangent at (4,4) and set its distance from the circle's centre equal to the radius.

Answer 1

The Correct Option is B

Step 1: Tangent by calculus
$2y\dfrac{dy}{dx}=4$, so the slope at $(4,4)$ is $\dfrac12$. The tangent is $y-4=\dfrac12(x-4)$, which is $x-2y+4=0$.

Step 2: Tangency condition
The perpendicular distance from the origin is $\dfrac{|4|}{\sqrt{1+4}}$, and $c$ is the square of this distance.

Step 3: Result
$c=\dfrac{16}5$, so $5c=16$.

Final Answer:
5c equals 16. \[ \boxed{\text{(B) }16} \]