The equation of the circle having centre at $(3, -4)$ and touching the line $5x + 12y - 12 = 0$ is ________.

The equation of the circle having centre at (3, -4 – Class 11 Mathematics

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(3, -4)

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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 equation of a circle with a given center and a tangent line is found by calculating the radius as the perpendicular distance from the center to the line, then substituting these values into the standard circle equation. The resulting equation is $\boxed{169x^2 + 169y^2 - 1014x + 1352y + 2200 = 0}$ .

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, $(x - h)^2 + (y - k)^2 = r^2$ , directly uses these, where $(h, k)$ is the center and $r$ is the radius.

The problem provides the center directly as $(3, -4)$ . This gives us $h=3$ and $k=-4$ .

The crucial piece of information is that the circle "touches the line $5x + 12y - 12 = 0$ ". When a circle touches a line, it means the line is tangent to the circle. A fundamental property of a tangent line is that the radius drawn to the point of tangency is perpendicular to the tangent line. This implies that the distance from the center of the circle to the tangent line is exactly equal to the radius of the circle.

Therefore, our strategy will be:

Identify the center $(h, k)$ .
Calculate the perpendicular distance from the center to the given tangent line. This distance will be the radius $r$ .
Substitute the center $(h, k)$ and the calculated radius $r$ into the standard equation of a circle.
Identify the Center

The problem states that the center of the circle is at $(3, -4)$ .

So, we have $h = 3$ and $k = -4$ .
Calculate the Radius

The radius $r$ is the perpendicular distance from the center $(h, k) = (3, -4)$ to the line $5x + 12y - 12 = 0$ .
$d$
In our case, $(x_1, y_1) = (3, -4)$ and the line is $5x + 12y - 12 = 0$ , which means $A=5$ , $B=12$ , and $C=-12$ .

Substituting these values into the distance formula:

$r = \frac{|5(3) + 12(-4) - 12|}{\sqrt{5^2 + 12^2}}$

$r = \frac{|15 - 48 - 12|}{\sqrt{25 + 144}}$

$r = \frac{|-45|}{\sqrt{169}}$

$r = \frac{45}{13}$

So, the radius of the circle is $\frac{45}{13}$ .
Formulate the Equation of the Circle …

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Q.The equation of the circle having centre at (3,−4)(3, -4)(3,−4) and touching the line 5x+12y−12=05x + 12y - 12 = 05x+12y−12=0 is ________.

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Q.The equation of the circle having centre at $(3, -4)$ and touching the line $5x + 12y - 12 = 0$ is ________.

Unlock everything free for 14 days