Question

The circle S = x² + y² − 2x − 4y + 1 = 0 cuts the y-axis at A, B (OA: OB). If the radical axis of S ≡ 0 and S′ = x² + y² − 4x − 2y + 4 = 0 cuts the y-axis at C, then the ratio in which C divides AB is:

• A. \(7 + 2\sqrt{3} : -7 + 2\sqrt{3}\)
• B. \(\sqrt{3} + 2 : \sqrt{3} - 2\)
• C. \(6 - 2\sqrt{3} : 2\sqrt{3} - 6\)
• D. \(-3 : \sqrt{3}\)

Hint:

Remember the formula for the radical axis and section formula.

Answer 1

The Correct Option is A

Step 1: Find the points A and B.
The circle S \equiv x² + y² - 2x - 4y + 1 = 0 cuts the y-axis at A and B.
For y-axis, put x = 0:
y² - 4y + 1 = 0
Using the quadratic formula, y = \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
y = \(\frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}\)
So, A = (0, 2 + √3) and B = (0, 2 - √3).

Step 2: Find the radical axis of the circles S and S'.
The radical axis of two circles S = 0 and S' = 0 is S - S' = 0.
S = x² + y² - 2x - 4y + 1 = 0
S' = x² + y² - 4x - 2y + 4 = 0
S - S' = (-2x - 4y + 1) - (-4x - 2y + 4) = 0
2x - 2y - 3 = 0

Step 3: Find the point C.
The radical axis cuts the y-axis at C.
Put x = 0 in 2x - 2y - 3 = 0:
-2y - 3 = 0
y = -\(\frac{3}{2}\)
So, C = (0, -\(\frac{3}{2}\)).

Step 4: Find the ratio in which C divides AB.
Let C divide AB in the ratio m : n.
Using section formula:
-\(\frac{3}{2}\) = \(\frac{m(2 - \sqrt{3}) + n(2 + \sqrt{3})}{m + n}\)
-\(\frac{3}{2}\)(m + n) = 2m - m√3 + 2n + n√3
-3m - 3n = 4m - 2m√3 + 4n + 2n√3
-7m - 7n = -2m√3 + 2n√3
-7(m + n) = 2√3(n - m)
-7m - 7n = 2√3n - 2√3m
(2√3 - 7)m = (2√3 + 7)n
\(\frac{m}{n}\) = \(\frac{2\sqrt{3} + 7}{2\sqrt{3} - 7} = \frac{7 + 2\sqrt{3}}{-7 + 2\sqrt{3}}\)
Therefore, the ratio is 7 + 2√3 : -7 + 2√3.