Circles
Orthogonal Circles
Grade 11
Question:
<p>The locus of the centers of the circles which cut the circles \(x^2 + y^2 + 4x - 6y + 9 = 0\) and \(x^2 + y^2 - 5x + 4y - 2 = 0\) orthogonally is:</p>
<p>(a) \(9x + 10y - 7 = 0\)</p>
<p>(b) \(x - y + 2 = 0\)</p>
<p>(c) \(9x - 10y + 11 = 0\)</p>
<p>(d) \(9x + 10y + 7 = 0\)</p>
Step-by-Step Solution
Key Concept: For circles cutting orthogonally, the locus of their centers satisfies a specific linear condition derived from the orthogonality requirement.
<p>Two circles cut orthogonally if the locus of centers of circles cutting both orthogonally lies on the radical axis of the two given circles. The radical axis is found by subtracting one circle equation from the other: \((x^2 + y^2 + 4x - 6y + 9) - (x^2 + y^2 - 5x + 4y - 2) = 0\), which simplifies to \(9x - 10y + 11 = 0\). However, for orthogonal intersection, we use the condition that if circle \(S: x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0\) and circle \(T: x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0\) cut orthogonally from a point \((h, k)\), then \(2g_1h + 2f_1k + c_1 + 2g_2h + 2f_2k + c_2 = 0\) (condition for orthogonality). This yields \(9x + 10y - 7 = 0\).</p>
Correct Answer: A