Question:
The centre of a circle \(S=0\) is at \((2,5)\) and its radius is \(r\). \(S_1=0\) is a circle which lies in the second quadrant and touches the coordinate axes and intersects the circle \(S=0\) at two points. If the radius of circle \(S_1=0\) is \(2\), then the possible values of \(r\) lie in the interval
The centre of a circle \(S=0\) is at \((2,5)\) and its radius is \(r\). \(S_1=0\) is a circle which lies in the second quadrant and touches the coordinate axes and intersects the circle \(S=0\) at two points. If the radius of circle \(S_1=0\) is \(2\), then the possible values of \(r\) lie in the interval
Show Hint
If two circles of radii \(r_1\) and \(r_2\) intersect at two distinct points, then
\[
|r_1-r_2|<d<r_1+r_2,
\]
where \(d\) is the distance between their centres.
Updated On: Jul 9, 2026
- \((2,8)\)
- \((8,14)\)
- \((3,7)\)
- \((7,10)\) \bigskip
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The Correct Option is C
Solution and Explanation
Concept:
A circle touching both coordinate axes in the second quadrant and having radius \(2\) must have centre
\[
(-2,2).
\]
Two circles intersect at two distinct points if the distance \(d\) between their centres satisfies
\[
|r_1-r_2|<d<r_1+r_2.
\]
Step 1: Find the centre of circle \(S_1\). Since \(S_1\) lies in the second quadrant and touches both coordinate axes, its centre is \[ (-2,2). \] Its radius is \[ 2. \]
Step 2: Find the distance between the centres. The centre of \(S\) is \[ (2,5). \] Hence, \[ d = \sqrt{(2+2)^2+(5-2)^2}. \] \[ = \sqrt{16+9}. \] \[ =5. \]
Step 3: Apply the condition for two-point intersection. Let the radius of \(S\) be \(r\). For two distinct points of intersection, \[ |r-2|<5<r+2. \] From \[ 5<r+2, \] \[ r>3. \] Also, \[ |r-2|<5. \] \[ -5<r-2<5. \] \[ -3<r<7. \] Since radius is positive, \[ r<7. \] Combining, \[ 3<r<7. \]
Step 4: Write the final answer. \[ \boxed{(3,7)} \]
Step 1: Find the centre of circle \(S_1\). Since \(S_1\) lies in the second quadrant and touches both coordinate axes, its centre is \[ (-2,2). \] Its radius is \[ 2. \]
Step 2: Find the distance between the centres. The centre of \(S\) is \[ (2,5). \] Hence, \[ d = \sqrt{(2+2)^2+(5-2)^2}. \] \[ = \sqrt{16+9}. \] \[ =5. \]
Step 3: Apply the condition for two-point intersection. Let the radius of \(S\) be \(r\). For two distinct points of intersection, \[ |r-2|<5<r+2. \] From \[ 5<r+2, \] \[ r>3. \] Also, \[ |r-2|<5. \] \[ -5<r-2<5. \] \[ -3<r<7. \] Since radius is positive, \[ r<7. \] Combining, \[ 3<r<7. \]
Step 4: Write the final answer. \[ \boxed{(3,7)} \]
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