Question

The equation of the circle touching the circle \[ x^2 + y^2 - 6x + 6y + 17 = 0 \] externally and to which the lines \[ x^2 - 3xy - 3x + 9y = 0 \] are normal is:

• A. \( x^2 + y^2 - 3x + 2y - 2 = 0 \)
• B. \( x^2 + y^2 - 6x - 2y + 1 = 0 \)
• C. \( x^2 + y^2 - 6x - 2y - 1 = 0 \)
• D. \( x^2 + y^2 - 9x - 3y + 2 = 0 \)

Hint:

When finding externally tangent circles, solve using the normal condition and ensure distance conditions are satisfied.

Answer 1

The Correct Option is B

Step 1: Finding the center and radius of the given circle Rewriting the given equation, \[ x^2 + y^2 - 6x + 6y + 17 = 0 \] Completing the square: \[ (x - 3)^2 - 9 + (y + 3)^2 - 9 + 17 = 0 \] \[ (x - 3)^2 + (y + 3)^2 = 1 \] So, the center is \( (3, -3) \) and radius \( R = 1 \). Step 2: Finding the required circle The required circle is externally tangent, meaning its center lies along the normal lines. Using the given normal line condition, we solve for the appropriate equation: \[ x^2 + y^2 - 6x - 2y + 1 = 0. \]