Question:
A circle has its centre in the first quadrant and passes through \((2,3)\). If this circle makes intercepts of length \(3\) and \(4\) respectively on \(x=2\) and \(y=3\), its equation is
A circle has its centre in the first quadrant and passes through \((2,3)\). If this circle makes intercepts of length \(3\) and \(4\) respectively on \(x=2\) and \(y=3\), its equation is
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For a circle, the chord length cut by a line at perpendicular distance \(d\) from the centre is \(2\sqrt{r^2-d^2}\).
Updated On: Jun 22, 2026
- \(x^2+y^2+3x-5y+8=0\)
- \(x^2+y^2-4x-6y+13=0\)
- \(x^2+y^2-6x-8y+23=0\)
- \(x^2+y^2-8x-9y+30=0\)
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The Correct Option is D
Solution and Explanation
Step 1: Assume the centre and radius of the circle.
Let the centre of the circle be \((h,k)\) and radius be \(r\).
Since the circle passes through \((2,3)\), we have
\[ r^2=(h-2)^2+(k-3)^2 \]
Step 2: Use the intercept on \(x=2\).
The line \(x=2\) is a vertical line.
The length of the chord made by the circle on \(x=2\) is \(3\).
The distance of the centre \((h,k)\) from \(x=2\) is
\[ |h-2| \] So, chord length is
\[ 2\sqrt{r^2-(h-2)^2}=3 \] Using \(r^2=(h-2)^2+(k-3)^2\), we get
\[ 2\sqrt{(k-3)^2}=3 \] \[ 2|k-3|=3 \] \[ |k-3|=\frac{3}{2} \]
Step 3: Use the intercept on \(y=3\).
The line \(y=3\) is a horizontal line.
The length of the chord made by the circle on \(y=3\) is \(4\).
The distance of the centre \((h,k)\) from \(y=3\) is
\[ |k-3| \] So, chord length is
\[ 2\sqrt{r^2-(k-3)^2}=4 \] Using \(r^2=(h-2)^2+(k-3)^2\), we get
\[ 2\sqrt{(h-2)^2}=4 \] \[ 2|h-2|=4 \] \[ |h-2|=2 \]
Step 4: Find the centre in the first quadrant.
Since the centre is in the first quadrant, we take
\[ h=2+2=4 \] and
\[ k=3+\frac{3}{2}=\frac{9}{2} \] Thus, the centre is
\[ \left(4,\frac{9}{2}\right) \]
Step 5: Find the radius.
\[ r^2=(h-2)^2+(k-3)^2 \] \[ r^2=(4-2)^2+\left(\frac{9}{2}-3\right)^2 \] \[ r^2=2^2+\left(\frac{3}{2}\right)^2 \] \[ r^2=4+\frac{9}{4} \] \[ r^2=\frac{25}{4} \]
Step 6: Write the equation of the circle.
The equation of the circle is
\[ (x-4)^2+\left(y-\frac{9}{2}\right)^2=\frac{25}{4} \] Expanding,
\[ x^2-8x+16+y^2-9y+\frac{81}{4}=\frac{25}{4} \] \[ x^2+y^2-8x-9y+16+\frac{81}{4}-\frac{25}{4}=0 \] \[ x^2+y^2-8x-9y+16+\frac{56}{4}=0 \] \[ x^2+y^2-8x-9y+30=0 \]
Step 7: Final conclusion.
Hence, the required equation of the circle is
\[ \boxed{x^2+y^2-8x-9y+30=0} \]
Let the centre of the circle be \((h,k)\) and radius be \(r\).
Since the circle passes through \((2,3)\), we have
\[ r^2=(h-2)^2+(k-3)^2 \]
Step 2: Use the intercept on \(x=2\).
The line \(x=2\) is a vertical line.
The length of the chord made by the circle on \(x=2\) is \(3\).
The distance of the centre \((h,k)\) from \(x=2\) is
\[ |h-2| \] So, chord length is
\[ 2\sqrt{r^2-(h-2)^2}=3 \] Using \(r^2=(h-2)^2+(k-3)^2\), we get
\[ 2\sqrt{(k-3)^2}=3 \] \[ 2|k-3|=3 \] \[ |k-3|=\frac{3}{2} \]
Step 3: Use the intercept on \(y=3\).
The line \(y=3\) is a horizontal line.
The length of the chord made by the circle on \(y=3\) is \(4\).
The distance of the centre \((h,k)\) from \(y=3\) is
\[ |k-3| \] So, chord length is
\[ 2\sqrt{r^2-(k-3)^2}=4 \] Using \(r^2=(h-2)^2+(k-3)^2\), we get
\[ 2\sqrt{(h-2)^2}=4 \] \[ 2|h-2|=4 \] \[ |h-2|=2 \]
Step 4: Find the centre in the first quadrant.
Since the centre is in the first quadrant, we take
\[ h=2+2=4 \] and
\[ k=3+\frac{3}{2}=\frac{9}{2} \] Thus, the centre is
\[ \left(4,\frac{9}{2}\right) \]
Step 5: Find the radius.
\[ r^2=(h-2)^2+(k-3)^2 \] \[ r^2=(4-2)^2+\left(\frac{9}{2}-3\right)^2 \] \[ r^2=2^2+\left(\frac{3}{2}\right)^2 \] \[ r^2=4+\frac{9}{4} \] \[ r^2=\frac{25}{4} \]
Step 6: Write the equation of the circle.
The equation of the circle is
\[ (x-4)^2+\left(y-\frac{9}{2}\right)^2=\frac{25}{4} \] Expanding,
\[ x^2-8x+16+y^2-9y+\frac{81}{4}=\frac{25}{4} \] \[ x^2+y^2-8x-9y+16+\frac{81}{4}-\frac{25}{4}=0 \] \[ x^2+y^2-8x-9y+16+\frac{56}{4}=0 \] \[ x^2+y^2-8x-9y+30=0 \]
Step 7: Final conclusion.
Hence, the required equation of the circle is
\[ \boxed{x^2+y^2-8x-9y+30=0} \]
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