Question

The centre of the circle whose radius is \( 5 \) and which touches the circle \( x^2 + y^2 - 2x - 4y - 20 = 0 \) at \( (5, 5) \) is:

• A. \( (10, 5) \)
• B. \( (5, 8) \)
• C. \( (5, 10) \)
• D. \( (8, 9) \)
• E. \( (9, 8) \)

Hint:

If two circles of the same radius touch externally, the point of contact is the midpoint of the line joining their centers.

Answer 1

Answer: (E)

Concept: When two circles touch each other at a point, their centers and the point of contact are collinear. The distance between the point of contact and the center of the new circle is equal to its radius. We can use the section formula or parametric coordinates of a line to find the new center.

Step 1: Find the center and radius of the given circle.
Equation: \( x^2 + y^2 - 2x - 4y - 20 = 0 \) Center \( C_1 = (1, 2) \). Radius \( R_1 = \sqrt{1^2 + 2^2 - (-20)} = \sqrt{25} = 5 \).

Step 2: Determine the direction of the center.
The point of contact is \( P(5, 5) \). The new circle has radius \( r = 5 \). Since the radius of the original circle is also 5, the point \( P(5, 5) \) is exactly midway between the two centers if they touch externally. Let the new center be \( C_2(h, k) \). Using the midpoint formula: \[ \frac{1 + h}{2} = 5 \quad \Rightarrow \quad 1 + h = 10 \quad \Rightarrow \quad h = 9 \] \[ \frac{2 + k}{2} = 5 \quad \Rightarrow \quad 2 + k = 10 \quad \Rightarrow \quad k = 8 \] Center \( C_2 = (9, 8) \).