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: \[ r = \sqrt{\left(\frac{PQ}{2}\right)^2 + d^2} = \sqrt{2^2 + (\sqrt{5})^2} = \sqrt{4 + 5} = \sqrt{9} = 3 \]

Answer 2

Step 1: Understand the problem.

We are given two circles:

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

The line joining two distinct points \( P \) and \( Q \) on Circle 1 is the diameter of Circle 2. This means:

• The distance between \( P \) and \( Q \) is equal to the diameter of Circle 2, which is \( 4 \) (since the radius is \( 2 \)).
• The midpoint of \( P \) and \( Q \) lies at the center of Circle 2, \( C_2(1, 3) \).

Step 2: Use the geometry of the problem.

The center of Circle 1 is \( C_1(2, 1) \), and the center of Circle 2 is \( C_2(1, 3) \). Since \( P \) and \( Q \) lie on Circle 1, their midpoint must coincide with \( C_2(1, 3) \). Therefore, \( C_2 \) lies on the line joining \( C_1 \) and the midpoint of \( P \) and \( Q \).

Step 3: Find the distance between \( C_1 \) and \( C_2 \).

The distance between \( C_1(2, 1) \) and \( C_2(1, 3) \) is:

\[ C_1C_2 = \sqrt{(2-1)^2 + (1-3)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}. \]

Step 4: Relate \( C_1C_2 \) to the radius \( r \).

Since \( P \) and \( Q \) are endpoints of a diameter of Circle 2, the distance \( C_1C_2 \) represents the perpendicular distance from \( C_1 \) to the chord (diameter) of Circle 2. This forms a right triangle with:

• Hypotenuse: \( r \) (radius of Circle 1),
• One leg: \( C_1C_2 = \sqrt{5} \),
• Other leg: Half the diameter of Circle 2, which is \( 2 \).

Using the Pythagorean theorem:

\[ r^2 = (\sqrt{5})^2 + 2^2 \]

\[ r^2 = 5 + 4 = 9 \]

\[ r = \sqrt{9} = 3 \]

Final Answer:

\( r = 3 \)