Question

The equation of the circle circumscribing the triangle whose sides are the lines y = x + 2, 3y = 4x and 2y = 3x, is __________.

Answer 1

For equations y = x + 2 (1)
3y = 4x (2)
2y = 3x (3)
Intersection point of (1) and (2) is given by
3(x + 2) = 4x
i.e. 3x + 6 = 4x
i.e. x = 6 and y = 8
Intersection point (2) and (3) is (0, 0) since both passes through origin.
Intersection point of (1) and (3) is given by
2(x + 2) = 3x
i.e. 2x + 4 = 3x
i.e. x = 4
y = 6
i.e. (4, 6) is the point intersection
Hence,
The co-ordinate of A, B and C are (4, 6), (6, 8) and (0, 0) respectively,
Let the equation of the circumcircle be x² + y² + 2ax + 2by + c = 0
Since the circle passes through A, B and C
i.e. 16 + 36 + 2a 4 + 2b 6 + c = 0
i.e. 52 + 8a + 12b + c = 0 …(4)
also, (6)² + (8)² + 12a + 16b + c = 0
i.e. 100 + 12a + 16b + c = 0 …(5)
and 0² + 0² + 2a × 0 + 2b × 0 + c = 0
i.e. c = 0 …(6)
∴ (4) and (5) reduces to,
$\left.\begin{array}{c}\begin{array}{lccccl}52+8a+12b=0& & & & & \mathrm{i}.\mathrm{e}.13+2\mathrm{a}+3\mathrm{b}=0\\ 100+12a+16b=0& & & & & \mathrm{i}.\mathrm{e}.25+3a+4b=0\end{array}\end{array}\right]\begin{array}{c}\times 3\\ \times 2\end{array}$
We get, 39 + 6a + 9b = 0
50 + 6a + 8b = 0
i.e. b = 11
and a = –23
∴ equation of circle is x² + y² – 46x + 22y = 0.