Question

A circle passes through the vertices of a quadrilateral ABCD. If the vertices are (5, 2), (–5, 2), (–5, –3) and (5, –3), then the radius of the circle is

• A. $3\sqrt{5}units$
• B. $\frac{3\sqrt{5}}{2}units$
  
• C. $\frac{5\sqrt{5}}{2}units$
  
• D. $5\sqrt{3}units$

Answer 1

The correct option is C $\frac{5\sqrt{5}}{2}units$


The coordinates of vertices of ABCD taken in order are A(5, 2) B(–5, 2), C(–5, –3) and D(5, –3).

Now, $AB=\sqrt{(5+5{)}^2+(2-2{)}^2}=10$



$BC=\sqrt{(-5+5{)}^2+(2+3{)}^2}=5$

$CD=\sqrt{(5+5{)}^2+(-3+3{)}^2}=10$

$AD=\sqrt{(5-5{)}^2+(2+3{)}^2}=5$

Thus, AB = CD and BC = AD.

Therefore, ABCD is a rectangle.

Now, the circle passes through A, B, C and D, so, its centre lies on the intersection of diagonal AC and BD.

$\therefore AC=\sqrt{(5+5{)}^2+(2+3{)}^2}=\sqrt{{10}^2+5^2}$

$=\sqrt{100+25}$

$=\sqrt{125}$

$=5\sqrt{5}$

$\therefore \text{Radius of circle}=\frac{1}{2}\times \text{Length of diagonal AC or BD}$

$=\frac{1}{2}\times 5\sqrt{5}$

$=\frac{5\sqrt{5}}{2}\text{units}$

Hence, the correct answer is option (c).