Find the equation of the circle which touches x-axis and whose centre is (1, 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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When a circle touches the $x$ -axis, its radius is the absolute value of the $y$ -coordinate of its center. Given the center $(1, 2)$ , the radius is $2$ , leading to the equation $(x-1)^2 + (y-2)^2 = 4$ .

To find the equation of a circle, we fundamentally need two pieces of information: its center and its radius. The standard form of a circle's equation directly uses these values.

The problem gives us the center directly. The crucial piece of information we need to interpret is "touches $x$ -axis" to find the radius.

Concept: Standard Form of a Circle and Touching an Axis

A circle is defined as the set of all points equidistant from a fixed point (the center). This constant distance is the radius.

If a circle has its center at $(h, k)$ and a radius $r$ , its equation is given by:

(h, k)

Now, consider what it means for a circle to "touch the $x$ -axis".

Imagine a circle with its center at $(h, k)$ . If this circle just touches the $x$ -axis, it means the $x$ -axis is tangent to the circle. The shortest distance from the center of the circle to the $x$ -axis must be equal to the radius.

The $x$ -axis is the line $y=0$ . The perpendicular distance from a point $(h, k)$ to the line $y=0$ is simply the absolute value of the $y$ -coordinate of the center, which is $|k|$ .

Therefore, if a circle touches the $x$ -axis, its radius $r$ must be equal to $|k|$ .

Tip

Visualizing this helps: If the center is $(1, 2)$ , the circle is in the first quadrant. For it to touch the $x$ -axis, its lowest point must be on the $x$ -axis. This means the vertical distance from the center $(1, 2)$ down to the $x$ -axis is the radius. This distance is $2$ units.

Let's apply this understanding to solve the problem.

Identify the given center:

The problem states that the center of the circle is $(1, 2)$ .

Comparing this with the standard form $(h, k)$ , we have $h=1$ and $k=2$ .
Determine the radius ( $r$ ):

The circle touches the $x$ -axis. As discussed, when a circle touches the $x$ -axis, its radius is the absolute value of the $y$ -coordinate of its center.

Here, the $y$ -coordinate of the center is $k=2$ . …

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Q.Find the equation of the circle which touches xxx-axis and whose centre is (1,2)(1, 2)(1,2).

Concept: Standard Form of a Circle and Touching an Axis

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Q.Find the equation of the circle which touches $x$ -axis and whose centre is $(1, 2)$ .

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