(iii)

The lines x + y = 2

3x – 4y = 6

x – y = 0

On solving these lines we get the intersection points A(2,0), B(- 6, – 6), C(1,1)

So by using the standard form of the equation of the circle:

x² + y² + 2ax + 2by + c = 0….. (1)

Substitute the points (2, 0) in equation (1), we get

\[\begin{array}{*{35}{l}}

{{2}^{2}}~+\text{ }{{0}^{2}}~+\text{ }2a\left( 2 \right)\text{ }+\text{ }2b\left( 0 \right)\text{ }+\text{ }c\text{ }=\text{ }0  \\

4\text{ }+\text{ }4a\text{ }+\text{ }c\text{ }=\text{ }0  \\

4a\text{ }+\text{ }c\text{ }+\text{ }4\text{ }=\text{ }0\ldots ..\text{ }\left( 2 \right)  \\

\end{array}\]
Substitute the point (-6, -6) in equation (1), we get

(- 6)² + (- 6)² + 2a(- 6) + 2b(- 6) + c = 0

36 + 36 – 12a – 12b + c = 0

12a + 12b – c – 72 = 0….. (3)

Substitute the points (1, 1) in equation (1), we get

\[\begin{array}{*{35}{l}}

{{1}^{2}}~+\text{ }{{1}^{2}}~+\text{ }2a\left( 1 \right)\text{ }+\text{ }2b\left( 1 \right)\text{ }+\text{ }c\text{ }=\text{ }0  \\

1\text{ }+\text{ }1\text{ }+\text{ }2a\text{ }+\text{ }2b\text{ }+\text{ }c\text{ }=\text{ }0  \\

2a\text{ }+\text{ }2b\text{ }+\text{ }c\text{ }+\text{ }2\text{ }=\text{ }0\ldots ..\text{ }\left( 4 \right)  \\

\end{array}\]
Upon simplifying equations (2), (3), (4) we get

a = 2, b = 3, c = – 12.

Substituting the values of a, b, c in equation (1), we get

\[\begin{array}{*{35}{l}}

{{x}^{2}}~+\text{ }{{y}^{2}}~+\text{ }2\left( 2 \right)x\text{ }+\text{ }2\left( 3 \right)y\text{ }-\text{ }12\text{ }=\text{ }0  \\

{{x}^{2}}~+\text{ }{{y}^{2}}~+\text{ }4x\text{ }+\text{ }6y\text{ }-\text{ }12\text{ }=\text{ }0  \\

\end{array}\]

∴ The equation of the circle is x² + y² + 4x + 6y – 12 = 0

(iv) y = x + 2, 3y = 4x and 2y = 3x

Given:

The lines \[\begin{array}{*{35}{l}}

y\text{ }=\text{ }x\text{ }+\text{ }2  \\

3y\text{ }=\text{ }4x  \\

2y\text{ }=\text{ }3x  \\

\end{array}\]

On solving these lines we get the intersection points A(6,8), B(0,0), C(4,6)

So by using the standard form of the equation of the circle:

x² + y² + 2ax + 2by + c = 0….. (1)

Substitute the points (6, 8) in equation (1), we get

\[\begin{array}{*{35}{l}}

{{6}^{2}}~+\text{ }{{8}^{2}}~+\text{ }2a\left( 6 \right)\text{ }+\text{ }2b\left( 8 \right)\text{ }+\text{ }c\text{ }=\text{ }0  \\

36\text{ }+\text{ }64\text{ }+\text{ }12a\text{ }+\text{ }16b\text{ }+\text{ }c\text{ }=\text{ }0  \\

12a\text{ }+\text{ }16b\text{ }+\text{ }c\text{ }+\text{ }100\text{ }=\text{ }0\ldots \ldots \text{ }\left( 2 \right)  \\

\end{array}\]
Substitute the points (0, 0) in equation (1), we get

\[\begin{array}{*{35}{l}}

{{0}^{2}}~+\text{ }{{0}^{2}}~+\text{ }2a\left( 0 \right)\text{ }+\text{ }2b\left( 0 \right)\text{ }+\text{ }c\text{ }=\text{ }0  \\

0\text{ }+\text{ }0\text{ }+\text{ }0a\text{ }+\text{ }0b\text{ }+\text{ }c\text{ }=\text{ }0  \\

c\text{ }=\text{ }0\ldots ..\text{ }\left( 3 \right)  \\

\end{array}\]
Substitute the points (4, 6) in equation (1), we get

\[\begin{array}{*{35}{l}}

{{4}^{2}}~+\text{ }{{6}^{2}}~+\text{ }2a\left( 4 \right)\text{ }+\text{ }2b\left( 6 \right)\text{ }+\text{ }c\text{ }=\text{ }0  \\

16\text{ }+\text{ }36\text{ }+\text{ }8a\text{ }+\text{ }12b\text{ }+\text{ }c\text{ }=\text{ }0  \\

8a\text{ }+\text{ }12b\text{ }+\text{ }c\text{ }+\text{ }52\text{ }=\text{ }0\ldots ..\text{ }\left( 4 \right)  \\

\end{array}\]
Upon simplifying equations (2), (3), (4) we get

a = – 23, b = 11, c = 0

Now by substituting the values of a, b, c in equation (1), we get

x² + y² + 2(- 23)x + 2(11)y + 0 = 0

x² + y² – 46x + 22y = 0

∴ The equation of the circle is x² + y² – 46x + 22y = 0