Question

The point on the circle \( (x-1)^2 + (y-1)^2 = 1 \) which is nearest to the circle \( (x-5)^2 + (y-5)^2 = 4 \) is

• A. \( (2,2) \)
• B. \( \left(\frac{3}{2}, \frac{3}{2}\right) \)
• C. \( \left(\frac{\sqrt{2}+1}{\sqrt{2}}, \frac{\sqrt{2}+1}{\sqrt{2}}\right) \)
• D. \( (\sqrt{2}, \sqrt{2}) \)
• E. \( (1+\sqrt{2}, 1+\sqrt{2}) \)

Hint:

When minimizing distance between geometric objects, always check the line joining their centers — it gives the direction of shortest distance.

Answer 1

The Correct Option is C

Concept:
• The shortest distance between two circles occurs along the line joining their centers.
• The nearest point on a circle from an external point lies along the line joining the center to that external point.
• Hence, the required point must lie on the line joining the centers of the two circles.

Step 1: Identify centers and radii of both circles.
First circle: \[ (x-1)^2 + (y-1)^2 = 1 \] Comparing with standard form \( (x-h)^2 + (y-k)^2 = r^2 \), we get: \[ C_1 = (1,1), \quad r_1 = 1 \] Second circle: \[ (x-5)^2 + (y-5)^2 = 4 \] \[ C_2 = (5,5), \quad r_2 = 2 \]

Step 2: Understanding geometrically what “nearest point” means.
We want the point on the first circle that is closest to the second circle. The shortest distance between two circles lies along the line joining their centers. Thus, the required point must lie on the straight line joining: \[ (1,1) \text{ and } (5,5) \]

Step 3: Equation of the line joining centers.
Slope: \[ m = \frac{5-1}{5-1} = 1 \] Equation: \[ y - 1 = 1(x - 1) \Rightarrow y = x \] So, required point lies on \( y = x \)

Step 4: Substitute into first circle.
Since \( y = x \), substitute into: \[ (x-1)^2 + (y-1)^2 = 1 \] \[ (x-1)^2 + (x-1)^2 = 1 \] \[ 2(x-1)^2 = 1 \]

Step 5: Solve for \(x\).
\[ (x-1)^2 = \frac{1}{2} \] \[ x-1 = \pm \frac{1}{\sqrt{2}} \] \[ x = 1 \pm \frac{1}{\sqrt{2}} \] Thus, two possible points: \[ \left(1 + \frac{1}{\sqrt{2}}, 1 + \frac{1}{\sqrt{2}}\right) \] \[ \left(1 - \frac{1}{\sqrt{2}}, 1 - \frac{1}{\sqrt{2}}\right) \]

Step 6: Choosing the nearest point.
The second circle lies in direction from \( (1,1) \) toward \( (5,5) \), i.e., increasing values. Thus, the nearer point is: \[ \left(1 + \frac{1}{\sqrt{2}}, 1 + \frac{1}{\sqrt{2}}\right) \]

Step 7: Final Answer.
\[ \boxed{\left( \frac{\sqrt{2}+1}{\sqrt{2}}, \frac{\sqrt{2}+1}{\sqrt{2}} \right)} \]