Question

Line \(l:x+y = 4\) intersects the circle \(x^2+y^2-2x-2y = 2\) at points \(A\) and \(B\). If C is the center of the circle, then the area of \(△ABC\) is...

• A. \(\sqrt{2}\)
• B. \(2\)
• C. \(2\sqrt{2}\)
• D. \(4\)

Hint:

Find the centre and radius, then the perpendicular distance from the centre to the line.

Answer 1

The Correct Option is B

Step 1: Centre and radius
\(x^2+y^2-2x-2y = 2\) becomes \((x-1)^2+(y-1)^2 = 4\). So \(C=(1,1)\) and \(r=2\).

Step 2: Distance from C to the line
\(d = \frac{|1+1-4|}{\sqrt2} = \sqrt2\).

Step 3: Chord length
Half the chord is \(\sqrt{r^2-d^2} = \sqrt{4-2} = \sqrt2\), so \(AB = 2\sqrt2\).

Step 4: Area
\[ \text{Area} = \frac12\times AB\times d = \frac12\times 2\sqrt2\times\sqrt2 = 2 \]

Final Answer:
The area of the triangle is 2 square units. \[ \boxed{\text{(B)}\ 2} \]