Circles
Orthogonal Circles
Grade 11
Question:
<p>P is a point \((a, b)\) in the first quadrant. If the two circles which pass through P and touch both the coordinate axes cut at right angles, then:</p>
<p>(a) \(a^2 - 6ab + b^2 = 0\)</p>
<p>(b) \(a^2 + 2ab - b^2 = 0\)</p>
<p>(c) \(a^2 - 4ab + b^2 = 0\)</p>
<p>(d) \(a^2 - 8ab + b^2 = 0\)</p>
Step-by-Step Solution
Key Concept: Find the radii of circles tangent to both axes passing through P, then apply orthogonality condition.
<p>A circle touching both coordinate axes in the first quadrant has center at \((r, r)\) for some radius r, and equation \((x-r)^2 + (y-r)^2 = r^2\). Since it passes through \((a,b)\): \((a-r)^2 + (b-r)^2 = r^2\), giving \(r^2 - 2(a+b)r + a^2 + b^2 = 0\). This yields two values \(r_1, r_2\) for the two circles. For perpendicular intersection (orthogonal circles), we use the condition: \(r_1^2 + r_2^2 = (\text{distance between centers})^2\). After substitution and simplification, we get \(a^2 - 8ab + b^2 = 0\).</p>
Correct Answer: D