Question

The radius of the least circle passing through the point (8,4) and cutting the circle x² + y² = 40 orthogonally is

(a)  √5

(b)  √7

(c)   2√5

(d)   4√5

Answer 1

Let the circle be x² + y² + 2gx + 2fy + c = 0 .......(1)

Given circle is x² + y² = 40 .......(2)

These two circles are orthogonal 

∴ c – 40 = 0

c = 40

(1) passes through (8,4)

64 + 16 + 16g + 8f + 40 = 0

120 + 16g + 8f = 0

f + 2g + 15 = 0

or f = –(2g + 15)

radius = √(g² + f² - c)

√(g² +(2g + 15)² - 40)

For least circle radius must be minimum

Let f(g) = g² + (2g + 15)² – 40 is minimum

f^'(g) = 2g + 4(2g + 15) = 0

10g = -60

g = -6

f''(g) = 10 > 0 minimum

f = –(–12 + 15) = –3

Equation of circle is x² + y² – 12x – 6y + 40 = 0

radius = √(36 + 9 - 40)

= √5