Question:
The equation of the circle passing through $(4, 5)$ and having the center at $(2, 2)$ is:
The equation of the circle passing through $(4, 5)$ and having the center at $(2, 2)$ is:
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The constant term in $x^2+y^2+2gx+2fy+c=0$ is $c = g^2+f^2-r^2$.
Updated On: Apr 8, 2026
- $x^{2} + y^{2} + 4x + 4y - 5 = 0$
- $x^{2} + y^{2} - 4x - 4y - 5 = 0$
- $x^{2} + y^{2} - 4x - 4y + 5 = 0$
- $x^{2} + y^{2} + 4x + 4y + 5 = 0$
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The Correct Option is B
Solution and Explanation
Step 1: Concept
Equation of a circle with center $(h, k)$ and radius $r$ is $(x-h)^{2} + (y-k)^{2} = r^{2}$.
Step 2: Analysis
Radius $r = \text{distance between } (2, 2) \text{ and } (4, 5) = \sqrt{(4-2)^{2} + (5-2)^{2}} = \sqrt{4+9} = \sqrt{13}$.
Equation: $(x-2)^{2} + (y-2)^{2} = (\sqrt{13})^{2} \Rightarrow x^{2} - 4x + 4 + y^{2} - 4y + 4 = 13$.
Step 3: Conclusion
$x^{2} + y^{2} - 4x - 4y - 5 = 0$.
Final Answer: (B)
Equation of a circle with center $(h, k)$ and radius $r$ is $(x-h)^{2} + (y-k)^{2} = r^{2}$.
Step 2: Analysis
Radius $r = \text{distance between } (2, 2) \text{ and } (4, 5) = \sqrt{(4-2)^{2} + (5-2)^{2}} = \sqrt{4+9} = \sqrt{13}$.
Equation: $(x-2)^{2} + (y-2)^{2} = (\sqrt{13})^{2} \Rightarrow x^{2} - 4x + 4 + y^{2} - 4y + 4 = 13$.
Step 3: Conclusion
$x^{2} + y^{2} - 4x - 4y - 5 = 0$.
Final Answer: (B)
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