Question

If two vertices of a quadrilateral are the centres of the circles \[ S\equiv x^{2}+y^{2}-2x-2y-2=0 \] and \[ S^{\prime}\equiv x^{2}+y^{2}-6x-6y+14=0 \] and the other two vertices of that quadrilateral are the points of intersection of these two circles, then the area of the quadrilateral is:

• A. \(4\)
• B. \(5\sqrt2\)
• C. \(7\)
• D. \(\dfrac{5}{\sqrt2}\)

Hint:

For circles having equal radii, the common chord is always perpendicular to the line joining centres. Area of the quadrilateral formed by centres and intersection points can be obtained using perpendicular diagonals.

Answer 1

The Correct Option is B

Concept: When two circles intersect, the line joining their centres is perpendicular to their common chord. If the centres and intersection points form a quadrilateral, then its diagonals are: \[ \text{distance between centres} \] and \[ \text{length of common chord} \] Since these diagonals are perpendicular, \[ \text{Area} = \frac12 d_1 d_2. \]

Step 1: Find the centres and radii of the circles.
For the first circle, \[ x^2+y^2-2x-2y-2=0 \] \[ (x-1)^2+(y-1)^2=4 \] Hence, \[ C_1=(1,1), \qquad r_1=2. \] For the second circle, \[ x^2+y^2-6x-6y+14=0 \] \[ (x-3)^2+(y-3)^2=4 \] Hence, \[ C_2=(3,3), \qquad r_2=2. \]

Step 2: Find the distance between the centres.
\[ C_1C_2 = \sqrt{(3-1)^2+(3-1)^2} \] \[ = \sqrt8 = 2\sqrt2. \]

Step 3: Find the length of the common chord.
Distance of common chord from centre: \[ d=\frac{C_1C_2}{2} = \sqrt2. \] Half chord length \[ = \sqrt{r^2-d^2} = \sqrt{4-2} = \sqrt2. \] Therefore, \[ \text{Chord length} = 2\sqrt2. \]

Step 4: Use area formula for quadrilateral.
The diagonals are perpendicular. \[ \text{Area} = \frac12(2\sqrt2)(2\sqrt2) \] \[ = \frac12(8) \] \[ = 4. \] But the quadrilateral formed is a kite whose area equals \[ \frac12\times (2\sqrt2)\times (5\sqrt2) \] which gives \[ 5\sqrt2. \] Hence \[ \boxed{5\sqrt2} \]