Question

Suppose the line joining distinct points P and Q on \((x-2)^{2}+(y-1)^{2}=r^{2}\) is the diameter of \( (x-1)^2+(y-3)^2=4\).The value of r is 

• A. \(3\)
• B. \(2\)
• C. \(1\)
• D. \(9\)
• E. \(4\)

Answer 1

The Correct Option is A

Step 1: Identify the Given Circles

- First Circle: \((x - 2)^2 + (y - 1)^2 = r^2\) with center \(C_1(2, 1)\) and radius \(r\).
- Second Circle: \((x - 1)^2 + (y - 3)^2 = 4\) with center \(C_2(1, 3)\) and radius \(2\) (since \(4 = 2^2\)).

Step 2: Understand the Condition

The line segment joining points \(P\) and \(Q\) on the first circle is the diameter of the second circle. This implies:

• The midpoint of \(P\) and \(Q\) is the center of the second circle, \(C_2(1, 3)\).
• The distance between \(C_1\) and \(C_2\) must satisfy a specific geometric relationship with \(r\).

 

Step 3: Calculate the Distance Between Centers

\[ \text{Distance} = \sqrt{(2 - 1)^2 + (1 - 3)^2} = \sqrt{1 + 4} = \sqrt{5} \]

Step 4: Determine the Radius \(r\)

Since \(PQ\) is the diameter of the second circle, its length is \(4\). The relationship between the distance between centers (\(d = \sqrt{5}\)) and the radius \(r\) is given by: \[ r = \sqrt{\left(\frac{PQ}{2}\right)^2 + d^2} = \sqrt{2^2 + (\sqrt{5})^2} = \sqrt{4 + 5} = \sqrt{9} = 3 \]