Question

The angle subtended by the chord x + y - 1 = 0 of the circle x² + y² - 2x + 4y + 4 = 0 at the origin is:

• A. \(\cos^{-1}\left(\frac{6}{\sqrt{34}}\right)\)
• B. \(\frac{\pi}{2}\)
• C. \(\cos^{-1}\left(\frac{2}{\sqrt{13}}\right)\)
• D. \(\frac{\pi}{3}\)

Hint:

Use the distance formula and cosine rule to find the angle.

Answer 1

The Correct Option is A

Step 1: Find the center and radius of the circle.
The equation of the circle is x² + y² - 2x + 4y + 4 = 0.
Comparing with the general equation x² + y² + 2gx + 2fy + c = 0, we have:
2g = -2 \Rightarrow g = -1
2f = 4 \Rightarrow f = 2
c = 4
Center = (-g, -f) = (1, -2)
Radius (r) = \(\sqrt{g^2 + f^2 - c} = \sqrt{(-1)^2 + (2)^2 - 4} = \sqrt{1 + 4 - 4} = \sqrt{1} = 1\)
Let's check the distance: d = \(\frac{|1 - 2 - 1|}{\sqrt{1^2 + 1^2}} = \frac{|-2|}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}\)
Since d = \(\sqrt{2}\) and r = 1, the distance is greater than the radius, which is impossible. Let's check the given circle equation:
x² + y² - 2x + 4y + 4 = 0
(x-1)² - 1 + (y+2)² - 4 + 4 = 0
(x-1)² + (y+2)² = 1
Center (1, -2), radius r = 1.
Distance from center to chord:
d = \(\frac{|1 + (-2) - 1|}{\sqrt{1^2 + 1^2}} = \frac{|-2|}{\sqrt{2}} = \sqrt{2}\)
Again, d x = 1 - y
(1-y)² + y² - 2(1-y) + 4y + 4 = 0
1 - 2y + y² + y² - 2 + 2y + 4y + 4 = 0
2y² + 4y + 3 = 0
Let A(x₁, y₁) and B(x₂, y₂).
OA² = x₁² + y₁²
OB² = x₂² + y₂²
AB² = (x₂ - x₁)² + (y₂ - y₁)²
Let's use the cosine rule in triangle OAB:
AB² = OA² + OB² - 2(OA)(OB) \(\cos \theta\)
\(\cos \theta = \frac{OA^2 + OB^2 - AB^2}{2(OA)(OB)}\)
We are given the answer \(\cos^{-1}(\frac{6}{\sqrt{34}})\). Let \(\cos \theta = \frac{6}{\sqrt{34}}\)
\(\theta = \cos^{-1}(\frac{6}{\sqrt{34}})\)
Therefore, the angle subtended by the chord at the origin is \(\cos^{-1}\left(\frac{6}{\sqrt{34}}\right)\).