Question

The point on a circle nearest to the point P(2, 1) is 4 units and the farthest point is (6, 5). Then the equation of the circle is –

• A. (x – 6) (x –$\sqrt { 2 }$– 1) + (y – 5) (y –$\sqrt { 2 }$– 1) = 0
• B. (x – 6) (x –$\sqrt { 2 }$) + (y – 5) (y –$\sqrt { 2 }$) = 0
• C. (x – 6) (x – 2 –2$\sqrt { 2 }$) + (y – 5) (y – 1 – 2$\sqrt { 2 }$) = 0
• D. (x – 6) (x – 2 –2$\sqrt { 2 }$) + (y – 5) (y – 2 – 2$\sqrt { 2 }$) = 0

Answer 1

Answer: (C)

(x – 6) (x – 2 –2$\sqrt { 2 }$) + (y – 5) (y – 1 – 2$\sqrt { 2 }$) = 0

Answer 2

We have

PB = $\sqrt { ( 6 - 2 ) ^ { 2 } + ( 5 - 1 ) ^ { 2 } }$ = $4 \sqrt { 2 }$

and PA = 4 (given)

Thus, AB = PB – PA = 4($\sqrt { 2 }$– 1)

Thus, $\frac { \sqrt { 2 } - 1 } { 1 }$

Hence, h = $\frac { 6 + 2 ( \sqrt { 2 } - 1 ) } { 1 + ( \sqrt { 2 } - 1 ) }$ = $\sqrt { 2 }$

and k = $\frac { 5 + ( \sqrt { 2 } - 1 ) } { 1 + ( \sqrt { 2 } - 1 ) }$ = $\frac { \sqrt { 2 } + 4 } { \sqrt { 2 } }$ = 1 + 2$\sqrt { 2 }$

The required circle has AB as its diameter.

Hence, its equation is

(x – 6) (x – 2 – 2$\sqrt { 2 }$) + (y – 5) (y – 1 – 2$\sqrt { 2 }$) = 0.