Question

The locus of the point of intersection of perpendicular tangents drawn to the circle \( x^2 + y^2 = 10 \) is:

• A. \( x^2 + y^2 = 5 \)
• B. \( x^2 + y^2 = 20 \)
• C. \( x^2 + y^2 = 25 \)
• D. \( x^2 + y^2 = 100 \)

Hint:

The locus of the intersection of perpendicular tangents to a circle of radius \( r \) follows the equation: \[ x^2 + y^2 = 2r^2. \]

Answer 1

The Correct Option is B

Step 1: Standard Equation of the Given Circle
The given circle equation is: \[ x^2 + y^2 = 10. \] This represents a circle centered at \( (0,0) \) with radius \( \sqrt{10} \).
Step 2: Equation of a Tangent to the Circle
The general equation of a tangent to a circle \( x^2 + y^2 = r^2 \) at a point \( (x_1, y_1) \) on the circle is: \[ x x_1 + y y_1 = r^2. \] For our given circle, the equation of the tangent at \( (x_1, y_1) \) is: \[ x x_1 + y y_1 = 10. \] Step 3: Condition for Perpendicular Tangents
If two tangents are perpendicular, their corresponding points of contact satisfy the property: \[ x_1 x_2 + y_1 y_2 = 0. \] Using the combined equation of two perpendicular tangents: \[ x^2 + y^2 = R^2 + r^2, \] where \( R \) is the radius of the locus circle and \( r \) is the radius of the given circle.
Step 4: Finding the Locus Equation
Using \( R^2 = 2r^2 \), we substitute \( r^2 = 10 \): \[ R^2 = 2(10) = 20. \] Thus, the required locus is: \[ x^2 + y^2 = 20. \] Step 5: Conclusion
Thus, the correct answer is \( \mathbf{x^2 + y^2 = 20} \).