Question

The least distance of the point A(10, 7) from the circle $x^2 + y^2 - 4x - 2y - 20 = 0$ is length of seg AM. If MM' is the diameter of the circle, then the lengths of AM and AM' are respectively ______, ______ units.

• A. 5, 10
• B. 5, 15
• C. 4, 15
• D. 2, 10

Hint:

From an external point to a circle:
Minimum Distance = (Distance to Center) - Radius.
Maximum Distance = (Distance to Center) + Radius.
These two extreme points always lie perfectly collinear with the center of the circle!

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Find the center $C$ and radius $r$ of the circle. The distance from point $A$ to the circle's boundary is related to the distance $AC$.

Step 2: Formula Application:
Circle: $(x-2)^2 + (y-1)^2 = 20 + 4 + 1 = 25$. Center $C(2, 1)$, Radius $r = 5$. Distance $AC = \sqrt{(10-2)^2 + (7-1)^2} = \sqrt{8^2 + 6^2} = 10$.

Step 3: Explanation:
The shortest distance $AM = AC - r = 10 - 5 = 5$ units. The longest distance $AM' = AC + r = 10 + 5 = 15$ units. Since $MM'$ is the diameter, $M$ is the closest point and $M'$ is the farthest point from $A$ along the line passing through the center.

Step 4: Final Answer:
The lengths are 5 and 15 units.