Question

Let a circle C of radius 1 and closer to the origin be such that the lines passing through the point (3, 2) and parallel to the coordinate axes touch it. Then the shortest distance of the circle C from the point (5, 5) is :

• A. \(2\sqrt2\)
• B. 5
• C. \(4\sqrt2\)
• D. 4

Answer 1

The Correct Option is D

Coordinates of the centre will be: \[ (2, 1) \]

Equation of circle: \[ (x - 2)^2 + (y - 1)^2 = 1 \]

 

\[ QC = \sqrt{(5 - 2)^2 + (5 - 1)^2} = 5 \]

Shortest distance: \[ RQ = CQ - CR = 5 - 1 = 4 \]

Answer 2

Step 1: Understand the tangency condition
Lines through (3, 2) parallel to the coordinate axes are the vertical line x = 3 and the horizontal line y = 2. If a circle of radius 1 is tangent to each of these lines, then the center (h, k) must satisfy \[ \text{dist}((h,k),\ x=3)=|h-3|=1,\qquad \text{dist}((h,k),\ y=2)=|k-2|=1. \] Hence \[ h\in\{2,4\},\qquad k\in\{1,3\}. \] So the possible centers are \[ (2,1),\ (2,3),\ (4,1),\ (4,3). \]

Step 2: Choose the circle closer to the origin
Compute squared distances to the origin to compare quickly: \[ (2,1):\ 2^2+1^2=5,\quad (2,3):\ 2^2+3^2=13,\quad (4,1):\ 4^2+1^2=17,\quad (4,3):\ 4^2+3^2=25. \] The smallest is 5, achieved at center (2, 1). Thus, the required circle C has center (2, 1) and radius r = 1.

Step 3: Shortest distance from the point (5, 5) to the circle
For a point P and a circle with center O and radius r, the shortest distance from P to the circle is \[ \max\{\,0,\ \overline{OP}-r\,\}=\overline{OP}-r\quad\text{if } \overline{OP}\ge r. \] Here, \(O=(2,1)\), \(P=(5,5)\), so \[ \overline{OP}=\sqrt{(5-2)^2+(5-1)^2}=\sqrt{3^2+4^2}=\sqrt{25}=5. \] Therefore, \[ \text{shortest distance}=5-1=4. \]

Final answer

4