Question:
The point of intersection of the lines
\[
2x+3y-12=0
\]
and
\[
3x-2y-5=0
\]
is the centre of a circle \(S=0\). If \(AB\) is a chord of \(S=0\) and it is a diameter of the circle
\[
x^2+y^2-10x+4y+13=0,
\]
then the radius of the circle \(S=0\) is
The point of intersection of the lines
\[
2x+3y-12=0
\]
and
\[
3x-2y-5=0
\]
is the centre of a circle \(S=0\). If \(AB\) is a chord of \(S=0\) and it is a diameter of the circle
\[
x^2+y^2-10x+4y+13=0,
\]
then the radius of the circle \(S=0\) is
Show Hint
If a chord of one circle is a diameter of another circle, first compute the diameter length. Then use
\[
\text{Chord Length}
=
2\sqrt{R^2-d^2}
\]
to determine the unknown radius.
Updated On: Jul 9, 2026
- \(4\)
- \(6\)
- \(16\)
- \(12\) \bigskip
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Concept:
If a chord of a circle is at a distance \(d\) from the centre and the circle has radius \(R\), then
\[
\text{Chord Length}
=
2\sqrt{R^2-d^2}.
\]
If the chord is a diameter of another circle, its length equals twice the radius of that circle.
Step 1: Find the centre of the circle \(S=0\). The centre is the point of intersection of \[ 2x+3y-12=0 \] and \[ 3x-2y-5=0. \] Solving, \[ 4x+6y=24, \] \[ 9x-6y=15. \] Adding, \[ 13x=39. \] \[ x=3. \] Substituting into \[ 3x-2y-5=0, \] \[ 9-2y-5=0. \] \[ y=2. \] Hence the centre of \(S=0\) is \[ C(3,2). \]
Step 2: Find the centre and radius of the given circle. Given \[ x^2+y^2-10x+4y+13=0. \] Completing squares, \[ (x-5)^2+(y+2)^2=16. \] Therefore, \[ C_1=(5,-2), \qquad r_1=4. \]
Step 3: Find the distance between the two centres. \[ CC_1 = \sqrt{(5-3)^2+(-2-2)^2}. \] \[ = \sqrt{4+16}. \] \[ = 2\sqrt5. \] Thus the distance from the centre of the given circle to the chord \(AB\) is \[ d=2\sqrt5. \]
Step 4: Find the length of the chord \(AB\). For the circle \[ (x-5)^2+(y+2)^2=16, \] radius \[ R=4. \] Hence, \[ AB = 2\sqrt{R^2-d^2}. \] \[ = 2\sqrt{16-20}. \] Since \(d>R\), the centre \(C(3,2)\) cannot represent the distance from the centre to the chord. Instead, \(AB\) is a diameter of the given circle. Therefore, \[ AB=2r_1=8. \]
Step 5: Use the chord-length formula for circle \(S\). The chord \(AB\) of circle \(S\) has midpoint at the centre of the given circle, \[ (5,-2). \] Distance from the centre of \(S\), \[ C(3,2), \] to the chord \(AB\) is \[ 2\sqrt5. \] Let the radius of \(S\) be \(R\). Then \[ 8 = 2\sqrt{R^2-20}. \] \[ 4 = \sqrt{R^2-20}. \] \[ R^2=36. \] \[ R=6. \]
Step 6: Write the final answer. \[ \boxed{6} \]
Step 1: Find the centre of the circle \(S=0\). The centre is the point of intersection of \[ 2x+3y-12=0 \] and \[ 3x-2y-5=0. \] Solving, \[ 4x+6y=24, \] \[ 9x-6y=15. \] Adding, \[ 13x=39. \] \[ x=3. \] Substituting into \[ 3x-2y-5=0, \] \[ 9-2y-5=0. \] \[ y=2. \] Hence the centre of \(S=0\) is \[ C(3,2). \]
Step 2: Find the centre and radius of the given circle. Given \[ x^2+y^2-10x+4y+13=0. \] Completing squares, \[ (x-5)^2+(y+2)^2=16. \] Therefore, \[ C_1=(5,-2), \qquad r_1=4. \]
Step 3: Find the distance between the two centres. \[ CC_1 = \sqrt{(5-3)^2+(-2-2)^2}. \] \[ = \sqrt{4+16}. \] \[ = 2\sqrt5. \] Thus the distance from the centre of the given circle to the chord \(AB\) is \[ d=2\sqrt5. \]
Step 4: Find the length of the chord \(AB\). For the circle \[ (x-5)^2+(y+2)^2=16, \] radius \[ R=4. \] Hence, \[ AB = 2\sqrt{R^2-d^2}. \] \[ = 2\sqrt{16-20}. \] Since \(d>R\), the centre \(C(3,2)\) cannot represent the distance from the centre to the chord. Instead, \(AB\) is a diameter of the given circle. Therefore, \[ AB=2r_1=8. \]
Step 5: Use the chord-length formula for circle \(S\). The chord \(AB\) of circle \(S\) has midpoint at the centre of the given circle, \[ (5,-2). \] Distance from the centre of \(S\), \[ C(3,2), \] to the chord \(AB\) is \[ 2\sqrt5. \] Let the radius of \(S\) be \(R\). Then \[ 8 = 2\sqrt{R^2-20}. \] \[ 4 = \sqrt{R^2-20}. \] \[ R^2=36. \] \[ R=6. \]
Step 6: Write the final answer. \[ \boxed{6} \]
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