If the line $lx + my = 1$ is a tangent to the circle $x^2 + y^2 = a^2$, then the point $(l, m)$ lies on a circle.

If the line lx + my = 1 is a tangent to the circle x² + y² = a², then the point (l, m) lies on a circle.

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lx + my = 1

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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 condition for tangency of a line to a circle gives a relation between $l$ and $m$ that is itself a circle. The point $(l, m)$ lies on $x^2 + y^2 = \frac{1}{a^2}$ .

Why this works

When a line touches a circle, the perpendicular distance from the circle's centre to the line equals the radius. That single geometric fact translates directly into an algebraic condition linking the line's coefficients. Here, the line is $lx + my = 1$ and the circle is $x^2 + y^2 = a^2$ — centred at the origin with radius $a$ .

The trick is to treat $l$ and $m$ not as fixed numbers but as coordinates of a point $(l, m)$ in a new plane. The tangency condition will produce an equation in $l$ and $m$ , and that equation turns out to be a circle.

Write the distance condition. For the line $lx + my - 1 = 0$ , the perpendicular distance from the centre $(0,0)$ to the line is

\frac{|l\cdot 0 + m\cdot 0 - 1|}{\sqrt{l^2 + m^2}} = \frac{1}{\sqrt{l^2 + m^2}}.

Since the line is tangent to the circle of radius $a$ , this distance must equal $a$ :

\frac{1}{\sqrt{l^2 + m^2}} = a.

Square both sides (both are positive, so no sign issues):

\frac{1}{l^2 + m^2} = a^2.

Rearrange to get an equation in $l$ and $m$ :

l^2 + m^2 = \frac{1}{a^2}.

Important

This is the equation of a circle centred at $(0,0)$ with radius $\frac{1}{a}$ .

Interpret the result. …

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Q.If the line lx+my=1lx + my = 1lx+my=1 is a tangent to the circle x2+y2=a2x^2 + y^2 = a^2x2+y2=a2, then the point (l,m)(l, m)(l,m) lies on a circle.

Why this works

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Q.If the line $lx + my = 1$ is a tangent to the circle $x^2 + y^2 = a^2$ , then the point $(l, m)$ lies on a circle.

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