Let the line x+y=1 meet the circle x^2+y^2=4 at the points A and B.

Question

Let the line $x+y=1$ meet the circle $x^2+y^2=4$ at the points $A$ and $B$. If the line perpendicular to $AB$ and passing through the midpoint of the chord $AB$ intersects the circle at $C$ and $D$, then the area of the quadrilateral $ACBD$ is equal to:

Accepted Answer

Foot of perpendicular from $O$ to $x+y=1$: $M=\left(\dfrac12,\dfrac12\right)$. The perpendicular through $M$ (slope $1$) is $y=x$, meeting the circle at $C=(\sqrt2,\sqrt2),\ D=(-\sqrt2,-\sqrt2)$, so $CD=4$.

Solving $x+y=1$ with $x^2+y^2=4$: $2x^2-2x-3=0\Rightarrow x=\dfrac{1\pm\sqrt7}{2}$, giving $A,B$ with

$$AB=\sqrt2\left|\frac{1+\sqrt7}{2}-\frac{1-\sqrt7}{2}\right|=\sqrt2\cdot\sqrt7=\sqrt{14}.$$

Since $AB\perp CD$ and they intersect (diagonals of $ACBD$):

$$\text{Area}=\frac12\cdot AB\cdot CD=\frac12(\sqrt{14})(4)=2\sqrt{14}.$$

Solutions · 1