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:
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- \( x^2 + y^2 - 3x + 2y - 2 = 0 \)
- \( x^2 + y^2 - 6x - 2y + 1 = 0 \)
- \( x^2 + y^2 - 6x - 2y - 1 = 0 \)
- \( x^2 + y^2 - 9x - 3y + 2 = 0 \)
The Correct Option is B
Solution and Explanation
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: \[ x^2 + y^2 - 6x - 2y + 1 = 0. \]
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