Question

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

• A. \(\frac{14π}{3}\)
• B. \(\frac{12π}{5}\)
• C. \(\frac{123π}{7}\)
• D. \(\frac{205π}{9}\)

Answer 1

The Correct Option is D

\(a=4,\; b=\sqrt{90},\; c=\sqrt{82}\)

Area of the triangle = \(\frac{1}{2}×4×9=18\)

The circumradius of the triangle \((R)=\frac{abc}{4Δ}\)

Area of the circle = \(πR^2 = π\bigg(\frac{abc}{4Δ}\bigg) = \frac{π(4.\sqrt{90}.\sqrt{82})^2}{(4.18)^2}\)

= \(π×\frac{16×90×82}{16×18×18} = \frac{205}{9}π\)

So, the correct answer is (D): \(\frac{205π}{9}\)

Answer 2

The equation of circle,
\((x - h)^2+ (y - k)^2 = r^2\)
All these points will satisfy this equation,

For \((0, 0)\) Put \(x = 0, y = 0\)
\(h^2 + k^2 = r^2 \)     \(........(1)\)

For \((4, 0)\) Put \(x = 4, y = 0\)
\((4- h)^2 + k^2= r^2\)      \(.......... (2)\)

For \((3, 9) \)Put \(x = 3, y = 9\)
\((3 - h)^2 + (9 - k)^2 = r^2\)     \(......... (3)\)

From \(eq(1)\) and \(eq(2)\),
\(h^2 + k^2= (4 - h)^2+ k^2\)
\(h^2= 16 +h^2-8h\)
\(8h = 16\)
\(h = 2, 0\)

From \(eq(3)- eq(1)\),
\(9 - 6h + 81 - 18k =0\)
\(90 - 18k = 6h\)
If \(h= 0\) then, \(k=5\)
If \(h = 2\) then, \(k = \frac {13}{3}\)
If \((h, k) = (0, 5)\)
\(r^2 = 25\)
\(\pi r^2 = 25\pi\)
If \((h, k) = (2, \frac {13}{3})\)

\(r^2 = 4 + \frac {169}{9}\)

\(r^2=\frac {205}{9}\)

\(\pi r^2=\frac {205\pi}{9}\)
Now, the area of circle \(=\frac {205\pi}{9}\)

So, the correct option is (D): \(\frac {205\pi}{9}\)