The point $(1, 2)$ lies inside the circle $x^2 + y^2 - 2x + 6y + 1 = 0$.

The point (1, 2) lies inside the circle x² + y² - – Class 11 Mathematics

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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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Substituting $(1,2)$ into $S(x,y)=x^2+y^2-2x+6y+1$ gives $S_1=16>0$ , so the point lies outside the circle. The statement is false.

For a circle $x^2+y^2+2gx+2fy+c=0$ and a point $P(x_1,y_1)$ , define $S_1=x_1^2+y_1^2+2gx_1+2fy_1+c$ . Then $P$ is inside if $S_1<0$ , on the circle if $S_1=0$ , and outside if $S_1>0$ .

Apply it. For the circle $x^2+y^2-2x+6y+1=0$ and point $(1,2)$ :

S_1=(1)^2+(2)^2-2(1)+6(2)+1=1+4-2+12+1=16

Since $S_1=16>0$ , the point is outside the circle. …

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Q.The point (1,2)(1, 2)(1,2) lies inside the circle x2+y2−2x+6y+1=0x^2 + y^2 - 2x + 6y + 1 = 0x2+y2−2x+6y+1=0.

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Q.The point $(1, 2)$ lies inside the circle $x^2 + y^2 - 2x + 6y + 1 = 0$ .

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