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

• A. \((2,8)\)
• B. \((8,14)\)
• C. \((3,7)\)
• D. \((7,10)\) \bigskip

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.

Answer 1

The Correct Option is C

Concept: For a circle in the second quadrant touching both axes, the centre is \((-r,r)\). Here \(r=2\), so centre = \((-2,2)\). Intersection condition: distance \(d\) between centres satisfies \(|r-2| < d < r+2\). Solve for \(r\).

Step 1: S1: centre (-2,2), radius 2. S: centre (2,5), radius r. Distance \(d = \sqrt{(2+2)^2+(5-2)^2} = \sqrt{16+9}=5\).

Step 2: Two distinct intersection points: \(|r-2| < 5 < r+2\). From \(5 < r+2 \Rightarrow r > 3\). From \(|r-2| < 5 \Rightarrow -5 < r-2 < 5 \Rightarrow -3 < r < 7\). With r>0, combine: \(3 < r < 7\).

Step 3: The range is \((3,7)\).

Step 4: Write the final answer. \(\boxed{(3,7)}\)