Question:
If the one end of a diameter of the circle $x^2 + y^2 + 3x + y - 6 = 0$ is at $(-4,-2)$, then the other end of the diameter is at
If the one end of a diameter of the circle $x^2 + y^2 + 3x + y - 6 = 0$ is at $(-4,-2)$, then the other end of the diameter is at
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Centre of circle = midpoint of diameter endpoints.
Updated On: Apr 24, 2026
- $(4,-2)$
- $(1,-1)$
- $(1,1)$
- $(-1,-1)$
- $(1,-2)$
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The Correct Option is C
Solution and Explanation
Concept:
• Centre of circle is midpoint of diameter
Step 1: Find centre
\[ x^2 + y^2 + 3x + y - 6 = 0 \] \[ \text{Centre} = \left(-\frac{3}{2}, -\frac{1}{2}\right) \]
Step 2: Use midpoint formula
Let other end be $(x,y)$ \[ \left(\frac{x-4}{2}, \frac{y-2}{2}\right) = \left(-\frac{3}{2}, -\frac{1}{2}\right) \]
Step 3: Solve equations
\[ \frac{x-4}{2} = -\frac{3}{2} \Rightarrow x = 1 \] \[ \frac{y-2}{2} = -\frac{1}{2} \Rightarrow y = 1 \] Final Conclusion:
\[ (1,1) \]
• Centre of circle is midpoint of diameter
Step 1: Find centre
\[ x^2 + y^2 + 3x + y - 6 = 0 \] \[ \text{Centre} = \left(-\frac{3}{2}, -\frac{1}{2}\right) \]
Step 2: Use midpoint formula
Let other end be $(x,y)$ \[ \left(\frac{x-4}{2}, \frac{y-2}{2}\right) = \left(-\frac{3}{2}, -\frac{1}{2}\right) \]
Step 3: Solve equations
\[ \frac{x-4}{2} = -\frac{3}{2} \Rightarrow x = 1 \] \[ \frac{y-2}{2} = -\frac{1}{2} \Rightarrow y = 1 \] Final Conclusion:
\[ (1,1) \]
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