This question is from Co-ordinate Geometry. Questions that appear in the CAT Exam from Co-ordinate Geometry tests an aspirant on the graphical representation Geometrical shapes, Distance between points, Section formula, Intercepts, Circles, and so on. In **CAT Exam**, one can generally expect to get approx. 1 question Coordinate Geometry. Make sure to a get good hold on this topic during your **CAT Preparation**. Practice **CAT Question Paper** from **2IIM CAT Question Bank** and get a kick-ass CAT score.

Question 19 : The vertices of a triangle are (0,0), (4,0) and (3,9). The area of the circle passing through these three points is

- \\frac{14π}{3})
- \\frac{123π}{7})
- \\frac{205π}{9})
- \\frac{12π}{5})

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A circle is circumscribed on the points (0,0) (4,0) and (3,9)

We are asked to find the area of this circle

We know that the area of a triangle with sides a, b and c, inscribed ina circle of radius R is given by

Area = \\frac{abc}{4R})

With the help of teh co-ordinates of the triangle it is easy to find the height of the triangle, which is QS = 9 units.

Hence the area of the triangle = A = \\frac{1}{2}) × base × height = \\frac{1}{2}) × PR × QS = \\frac{1}{2}) × 4 × 9 = 18 sq. units

a = PR = 4 units

b = QR = √(QS^{2} + SR^{2}) = √(9^{2} + 1^{2}) = √82

c = PQ = √(PS^{2} + PQ^{2}) = √(3^{2} + 9^{2}) = √90

W.K.T Area = \\frac{abc}{4R})

18 = \\frac{4 × √82 × √90}{4R})

R = \\frac{√205}{3})

Now that we know the radius of the circle the area of the circle = π × R^{2} = π × \\frac{√205}{3}) × \\frac{√205}{3}) = π × \\frac{205}{9}) = \\frac{205π}{9})

The question is **"The vertices of a triangle are (0,0), (4,0) and (3,9). The area of the circle passing through these three points is" **

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