Question

If the pair of tangents drawn to the circle \( x^2 + y^2 = a^2 \) from the point \( (10, 4) \) are perpendicular, then \( a \) is:

• A. \( \sqrt{58} \)
• B. \( 58 \)
• C. \( 2\sqrt{63} \)
• D. \( 2\sqrt{45} \)

Hint:

For perpendicular tangents, use the property \( h^2 + k^2 = 2a^2 \).

Answer 1

The Correct Option is A

To find the value of \(a\) for which the pair of tangents drawn from the point \((10, 4)\) to the circle \(x^2+y^2=a^2\) are perpendicular, follow these steps:

The general equation of the pair of tangents from point \( (x_1, y_1) \) to the circle \( x^2 + y^2 = a^2 \) is given by:

\((x_1^2 + y_1^2 - a^2)(x^2 + y^2) = (x_1x + y_1y)^2\)

Substitute \(x_1 = 10\) and \(y_1 = 4\):

\((10^2+4^2-a^2)(x^2+y^2) = (10x+4y)^2\)

Simplify:

\((100 + 16 - a^2)(x^2 + y^2) = (10x + 4y)^2\)

\((116 - a^2)(x^2 + y^2) = 100x^2 + 80xy + 16y^2\)

Tangents from \((10, 4)\) are perpendicular if:

\(c = 0\) where \(c\) is the constant in the homogeneous form \(ax^2 + 2hxy + by^2\) with \((a + b = 0)\).

The equation is a homogeneous quadratic:

\(100x^2 + 80xy + 16y^2 - (116-a^2)(x^2+y^2) = 0\)

The equation becomes:

\((100-(116-a^2))x^2 + 80xy + ((16-(116-a^2))y^2) = 0\)

We equate terms for zero sum:

\(100-(116-a^2) + 16-(116-a^2) = 0\)

\(100 + 16 - 232 + 2a^2 = 0\)

\(116 - 232 + 2a^2 = 0\)

\(2a^2 = 116\)

\(a^2 = 58\)

\(a = \sqrt{58}\)

Thus, the value of \(a\) is \(\sqrt{58}\).

Answer 2

Step 1: Understanding the Condition for Perpendicular Tangents The given equation of the circle is: \[ x^2 + y^2 = a^2 \] A pair of tangents drawn from an external point \( (x_1, y_1) \) to a circle are perpendicular if and only if the given point lies on the director circle of the given circle.
Step 2: Equation of the Director Circle The equation of the director circle of a given circle \( x^2 + y^2 = a^2 \) is given by: \[ x^2 + y^2 = 2a^2 \] Since the point \( (10,4) \) lies on the director circle, we substitute \( x = 10 \) and \( y = 4 \): \[ 10^2 + 4^2 = 2a^2 \] \[ 100 + 16 = 2a^2 \]
Step 3: Solving for \( a \) Dividing by 2: \[ a^2 = 58 \] \[ a = \sqrt{58} \]
Final Answer: \[ \boxed{\sqrt{58}} \]