The equation of the circle circumscribing the triangle whose sides are the lines y = x + 2, 3y = 4x, 2y = 3x i

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y = x + 2

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Concept understanding — Circle Equation Standard Form

Where the Circle Equation Comes From

Imagine you're standing at a point on a flat field. You tie a rope to a stake at that point, walk out the full length of the rope, and start walking in a circle, keeping the rope taut. Every point you step on is exactly the same distance from the stake.

That's the entire idea: a circle is the set of all points that are a fixed distance (the radius) from a fixed point (the centre).

If we put this on a coordinate plane, we can turn that geometric idea into an algebraic equation.

From Geometry to Algebra

Let the centre be at coordinates $(h, k)$ . Let the radius be $r$ . Take any point $(x, y)$ that lies on the circle. The distance from $(x, y)$ to $(h, k)$ must equal $r$ .

What's the distance between two points in the plane? The distance formula:

\sqrt{(x - h)^2 + (y - k)^2} = r

Now square both sides to remove the square root:

(x - h)^2 + (y - k)^2 = r^2

That's it. That's the standard form of the equation of a circle.

(x - h)^2 + (y - k)^2 = r^2

What Each Piece Tells You

$(x - h)$ and $(y - k)$ — these shift the circle away from the origin. If the centre is at $(0,0)$ , the equation simplifies to $x^2 + y^2 = r^2$ .
$r^2$ — notice it's the square of the radius, not the radius itself. If the equation says $x^2 + y^2 = 25$ , the radius is $\sqrt{25} = 5$ , not 25.
The equals sign — only the points $(x, y)$ that make this equation true lie on the circle. Any other point gives a larger or smaller left-hand side.

Watch out

A common mistake: for $(x - 3)^2 + (y + 2)^2 = 16$ , students often read the centre straight off the signs printed in the equation and say $(3, 2)$ . That's wrong. Each bracket must first be written in the exact form $x - h$ and $y - k$ : here $(y + 2) = (y - (-2))$ , so $k = -2$ , not $2$ . The centre is actually $(3, -2)$ . Always flip the sign inside every bracket before reading off $h$ and $k$ .

Quick Example

Write the equation of a circle with centre $(-1, 4)$ and radius $3$ .

Here $h = -1$ , $k = 4$ , $r = 3$ . Plug in:

(x - (-1))^2 + (y - 4)^2 = 3^2

Simplify:

$(x + 1)^2 + (y - 4)^2 = 9$ …

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The circumcircle of a triangle can be found by first solving the three line equations pairwise to get the vertices, then substituting those coordinates into the general circle equation $x^2 + y^2 + 2gx + 2fy + c = 0$ and solving for $g$ , $f$ , $c$ . The final equation is $x^2 + y^2 - 46x + 22y = 0$ .

We are given three lines:

$L_1: y = x + 2$

$L_2: 3y = 4x$

$L_3: 2y = 3x$

These form a triangle. The circle that passes through all three vertices is the circumcircle. The most direct method: find the vertices (intersection points), then find the circle through them.

1. Find the vertices of the triangle

Solve each pair of lines.

Vertex A: intersection of $L_1$ and $L_2$

$y = x + 2$ and $3y = 4x$

Substitute $y$ : $3(x + 2) = 4x \implies 3x + 6 = 4x \implies x = 6$

Then $y = 6 + 2 = 8$

So $A = (6, 8)$

Vertex B: intersection of $L_1$ and $L_3$

$y = x + 2$ and $2y = 3x$

Substitute: $2(x + 2) = 3x \implies 2x + 4 = 3x \implies x = 4$

Then $y = 4 + 2 = 6$

So $B = (4, 6)$

Vertex C: intersection of $L_2$ and $L_3$

$3y = 4x$ and $2y = 3x$

From $3y = 4x$ , we have $y = \frac{4x}{3}$ . Substitute into $2y = 3x$ :

$2 \cdot \frac{4x}{3} = 3x \implies \frac{8x}{3} = 3x \implies 8x = 9x \implies x = 0$

Then $y = 0$

So $C = (0, 0)$

Tip

Notice that $C = (0,0)$ is the origin. That simplifies the circle equation because the constant term $c$ will be zero if the origin lies on the circle — but we must verify. Actually, if $(0,0)$ is on the circle, then plugging into $x^2 + y^2 + 2gx + 2fy + c = 0$ gives $c = 0$ . So we already know $c = 0$ .

2. Set up the general circle equation

The general circle:

$x^2 + y^2 + 2gx + 2fy + c = 0$

Since $C = (0,0)$ lies on it:

$0 + 0 + 0 + 0 + c = 0 \implies c = 0$

So the equation reduces to:

$x^2 + y^2 + 2gx + 2fy = 0$

3. Plug in the other two vertices

For $A = (6, 8)$ :

$6^2 + 8^2 + 2g(6) + 2f(8) = 0$

$36 + 64 + 12g + 16f = 0$

$100 + 12g + 16f = 0$

Divide by 2: $50 + 6g + 8f = 0$ … (1)

For $B = (4, 6)$ :

$4^2 + 6^2 + 2g(4) + 2f(6) = 0$

$16 + 36 + 8g + 12f = 0$

$52 + 8g + 12f = 0$ …

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Q.The equation of the circle circumscribing the triangle whose sides are the lines y=x+2y = x + 2y=x+2, 3y=4x3y = 4x3y=4x, 2y=3x2y = 3x2y=3x is ________.

1. Find the vertices of the triangle

2. Set up the general circle equation

3. Plug in the other two vertices

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

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