Question:
Suppose that the \(x\)-coordinates of the points \(A\) and \(B\) satisfy
\[
x^2+2x-a^2=0
\]
and their \(y\)-coordinates satisfy
\[
y^2+4y-b^2=0.
\]
Then the equation of the circle with \(AB\) as its diameter is
Suppose that the \(x\)-coordinates of the points \(A\) and \(B\) satisfy
\[
x^2+2x-a^2=0
\]
and their \(y\)-coordinates satisfy
\[
y^2+4y-b^2=0.
\]
Then the equation of the circle with \(AB\) as its diameter is
Show Hint
If \(A(x_1,y_1)\) and \(B(x_2,y_2)\) are endpoints of a diameter, then the circle is:
\[
(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0.
\]
Updated On: Jun 22, 2026
- \(x^2+y^2+2x+4y-a^2-b^2=0\)
- \(x^2+y^2+2x+4y+a^2+b^2=0\)
- \(x^2+y^2-2x-4y-a^2-b^2=0\)
- \(x^2+y^2-2x-4y+a^2+b^2=0\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Let the endpoints of diameter be \(A(x_1,y_1)\) and \(B(x_2,y_2)\).
The \(x\)-coordinates satisfy \[ x^2+2x-a^2=0 \] So, \[ x_1+x_2=-2 \] and \[ x_1x_2=-a^2 \] The \(y\)-coordinates satisfy \[ y^2+4y-b^2=0 \] So, \[ y_1+y_2=-4 \] and \[ y_1y_2=-b^2 \]
Step 2: Use equation of circle with diameter endpoints.
The circle with \(AB\) as diameter is \[ (x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0 \]
Step 3: Expand the equation.
\[ x^2-(x_1+x_2)x+x_1x_2+y^2-(y_1+y_2)y+y_1y_2=0 \]
Step 4: Substitute values.
\[ x^2-(-2)x-a^2+y^2-(-4)y-b^2=0 \] \[ x^2+y^2+2x+4y-a^2-b^2=0 \]
Step 5: Final conclusion.
Therefore, \[ \boxed{x^2+y^2+2x+4y-a^2-b^2=0} \]
The \(x\)-coordinates satisfy \[ x^2+2x-a^2=0 \] So, \[ x_1+x_2=-2 \] and \[ x_1x_2=-a^2 \] The \(y\)-coordinates satisfy \[ y^2+4y-b^2=0 \] So, \[ y_1+y_2=-4 \] and \[ y_1y_2=-b^2 \]
Step 2: Use equation of circle with diameter endpoints.
The circle with \(AB\) as diameter is \[ (x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0 \]
Step 3: Expand the equation.
\[ x^2-(x_1+x_2)x+x_1x_2+y^2-(y_1+y_2)y+y_1y_2=0 \]
Step 4: Substitute values.
\[ x^2-(-2)x-a^2+y^2-(-4)y-b^2=0 \] \[ x^2+y^2+2x+4y-a^2-b^2=0 \]
Step 5: Final conclusion.
Therefore, \[ \boxed{x^2+y^2+2x+4y-a^2-b^2=0} \]
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