Question

If the tangent at $(1, 7)$ 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. 195
• B. 185
• C. 85
• D. 95

Answer 1

The Correct Option is D

To find the value of \(c\) such that the tangent at the point \((1, 7)\) to the curve \(x^2 = y - 6\) touches the circle \(x^2 + y^2 + 16x + 12y + c = 0\), we follow these steps:

First, we need to determine the equation of the tangent to the curve at the given point \((1, 7)\).

The curve is \(x^2 = y - 6\), or equivalently, \(y = x^2 + 6\).

Differentiating both sides with respect to \(x\) gives:

\(\frac{dy}{dx} = 2x\)

At the point \((1, 7)\), the slope of the tangent is \(2 \times 1 = 2\).

The equation of the tangent line at \((1, 7)\) is:

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

Rearranging, we get the tangent line equation:

\(y = 2x + 5\)

Next, we'll verify that this tangent line touches the circle \(x^2 + y^2 + 16x + 12y + c = 0\).

Substituting \(y = 2x + 5\) into the circle's equation:

\(x^2 + (2x + 5)^2 + 16x + 12(2x + 5) + c = 0\)

Expanding \((2x + 5)^2\) gives:

\(4x^2 + 20x + 25\)

Substituting gives:

\(x^2 + 4x^2 + 20x + 25 + 16x + 24x + 60 + c = 0\)

Simplifying:

\(5x^2 + 60x + 85 + c = 0\)

For the line to be tangent to the circle, the discriminant of this quadratic equation in \(x\) must be zero:

\(b^2 - 4ac = 0\)

Here, \(a = 5\), \(b = 60\), and \(c = 85 + c\).

Calculating the discriminant:

\(60^2 - 4 \times 5 \times (85 + c) = 0\)

\(3600 - 20(85 + c) = 0\)

\(3600 - 1700 - 20c = 0\)

Simplifying gives:

\(1900 = 20c\)

Solving for \(c\) gives:

\(c = 95\)

Therefore, the value of \(c\) such that the tangent touches the circle is 95.