Question:
A circle has radius 3 and its center lies on the line \( y = x - 1 \). The equation of the circle, if it passes through (7, 3), is:
A circle has radius 3 and its center lies on the line \( y = x - 1 \). The equation of the circle, if it passes through (7, 3), is:
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When the center of a circle lies on a line, you can use the distance formula and the equation of the circle to find its exact equation.
Updated On: Jan 12, 2026
- \( x^2 + y^2 - 8x - 6y + 16 = 0 \)
- \( x^2 + y^2 - 8x + 6y + 16 = 0 \)
- \( x^2 + y^2 - 8x - 6y - 16 = 0 \)
- \( x^2 + y^2 - 8x + 6y - 16 = 0 \)
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The Correct Option is A
Solution and Explanation
Using the center-radius form of the equation of a circle and the given conditions, we derive the equation as \( x^2 + y^2 - 8x - 6y + 16 = 0 \).
Final Answer: \[ \boxed{x^2 + y^2 - 8x - 6y + 16 = 0} \]
Final Answer: \[ \boxed{x^2 + y^2 - 8x - 6y + 16 = 0} \]
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