Question

Let f(x, y) = 0 represents a circle, If f(0, a) has roots a = 2, 3 and f(a, 0) has roots \(a = 12, \frac 1 2,\) then the centre of the circle is:
1. \(\left(\frac {-5}2, \frac 1 6\right)\)
2. \(\left(\frac {4}{25}, \frac 2 5\right)\)
3. \(\left(\frac {2}{7}, 6\right)\)
4. \(\left(\frac {25}{4}, \frac 5 2\right)\)

Answer 1

Correct Answer - Option 4 : \(\left(\frac {25}{4}, \frac 5 2\right)\)

f(x, y) = 0 → circle

Let center of the circle is (-f, -g) and

f(x, y) = x² + y² + 2fx + 2gy + c = 0

Given:

f(0, a) = a² + 2ag + c = 0

roots a = 2, 3

∴ 4 + 4g + c = 0     …1)

9 + 6g + c = 0     …2)

Solving 1) and 2) we get,

9 + 6g – 4g – 4 = 0

2g = -5

g = -2.5

Now,

f(a, 0) = a² + 2fa + c = 0

roots a = 12, 0.5

∴ 144 + 24f + c = 0     …3)

0.25 + f + c = 0     …4)

Solving 3) and 4) we get,

-143.75 = 23f

f = -6.25

The center is \(\left( {\frac{{25}}{4},\;\frac{5}{2}} \right)\)