Question:medium

The equation of mirror image of the circle \(x^2 + y^2 - 6x - 10y + 33 = 0\) about the line \(y = x\) is:

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Reflection in \(y=x\) $\Rightarrow$ swap \(x\) and \(y\).
Updated On: Apr 17, 2026
  • \(x^2 + y^2 - 10x + 6y + 33 = 0\)
  • \(x^2 + y^2 + 10x - 6y + 33 = 0\)
  • \(x^2 + y^2 - 10x - 6y + 33 = 0\)
  • \(x^2 + y^2 + 10x + 6y + 33 = 0\)
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The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:
The mirror image of a circle is another circle with the same radius, but its center is the mirror image of the original center. For reflection in \(y = x\), we simply swap the \(x\) and \(y\) coordinates.
Step 2: Detailed Explanation:
1. Original circle: \(x^{2} + y^{2} - 6x - 10y + 33 = 0\).
Center \(C = (3, 5)\).
Radius \(r = \sqrt{g^{2} + f^{2} - c} = \sqrt{3^{2} + 5^{2} - 33} = \sqrt{9 + 25 - 33} = 1\).
2. Image of center \(C(3, 5)\) about \(y = x\) is \(C'(5, 3)\).
3. Equation of new circle with center \((5, 3)\) and radius \(r = 1\):
\[ (x - 5)^{2} + (y - 3)^{2} = 1^{2} \]
\[ x^{2} - 10x + 25 + y^{2} - 6y + 9 = 1 \]
\[ x^{2} + y^{2} - 10x - 6y + 33 = 0 \]
Step 3: Final Answer:
The equation is \(x^{2} + y^{2} - 10x - 6y + 33 = 0\).
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