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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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 19 on this concept.

KEAM 2026Set eng-2026-04174 marksMCQ
$x^2 + y^2 + 8x - 6y + c = 0$
›Reveal solutionSolution

For $x^2+y^2+8x-6y+c=0$ , $r^2=g^2+f^2-c$ ; set $\pi r^2=75\pi$ .

Here $2g=8,\,2f=-6$ , so $g=4,\,f=-3$ and $r^2=g^2+f^2-c=16+9-c=25-c$ . …
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KEAM 2026Set eng-2026-04194 marksMCQ
$x^{2} + y^{2} + 3x + y - 6 = 0$
›Reveal solutionSolution

The centre bisects any diameter, so the other endpoint is $2C-P_1$ .

For $x^2+y^2+3x+y-6=0$ , the centre is $C=(-\tfrac32,-\tfrac12)$ .

With one end $P_1=(-4,-2)$ , the other end is …
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KEAM 2026Set eng-2026-04204 marksMCQ
$x^2+y^2=25$
›Reveal solutionSolution

The endpoints of the diameter are the intersections of $x+y=5$ with $x^2+y^2=25$ , namely $(0,5)$ and $(5,0)$ . The circle with this diameter is $(x-0)(x-5)+(y-5)(y-0)=0$ , i.e. $x^2+y^2-5x-5y=0$ .

Find where the chord $x+y-5=0$ meets the circle $x^2+y^2=25$ . Put $y=5-x$ :
$x^2+(5-x)^2=25\;\Rightarrow\;2x^2-10x=0\;\Rightarrow\;2x(x-5)=0,$
so $x=0$ (then $y=5$ ) or $x=5$ (then $y=0$ ). The endpoints are $(0,5)$ and $(5,0)$ . …
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KEAM 2026Set eng-2026-04214 marksMCQ
$x^{2}+y^{2}-2x+4y=8$
›Reveal solutionSolution

Completing the square gives centre $(1,-2)$ and radius $\sqrt{13}$ .

For $x^2+y^2-2gx-2fy+c=0$ the centre is $(g,f)$ and radius $\sqrt{g^2+f^2-c}$ . Here $x^2+y^2-2x+4y-8=0$ gives $g=1,\ f=-2,\ c=-8$ : …
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KEAM 2025Set eng-2025-04234 marksMCQ
$x$
›Reveal solutionSolution
The circle is $(x-9)^2+(y-7)^2=49$ .

Concept and Intuition

A circle tangent to the $x$ -axis at $(9,0)$ has its centre vertically above that point at $(9,r)$ . Tangency to a horizontal line above fixes $r$ .

Step-by-Step Solution

Centre $=(9,r)$ , radius $r$ (touches $x$ -axis at $(9,0)$ ).

Touches $y=14$ : distance $|14-r|=r\Rightarrow 14-r=r\Rightarrow r=7$ . …
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KEAM 2025Set eng-2025-04254 marksMCQ
$(-4,2)$
›Reveal solutionSolution

Find the circle's centre, then the line through it and the given point; both share $y=2$ , so the line is $y=2$ .

Divide the circle equation by 2: $x^2 + y^2 - 4y = \tfrac72$ . Complete the square: $x^2 + (y-2)^2 = \tfrac72 + 4$ . The centre is $(0,2)$ . …
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KEAM 2025Set eng-2025-04274 marksMCQ
$4x^2+24x+9y^2-18y+9=0$
›Reveal solutionSolution

Complete the square in $x$ and $y$ ; the ellipse becomes $4(x+3)^2+9(y-1)^2=36$ , centre $(-3,1)$ .

Start from $4x^2+24x+9y^2-18y+9=0$ :
$4(x^2+6x)+9(y^2-2y)+9=0.$
$4(x+3)^2-36+9(y-1)^2-9+9=0.$ …
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KEAM 2025Set eng-2025-04294 marksMCQ
$(4,0)$
›Reveal solutionSolution

Centre $(0,k)$ equidistant from $(4,0)$ and $(0,2)$ : $16+k^2=(k-2)^2\Rightarrow k=-3$ ; radius $=\sqrt{16+9}=5$ . …

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KEAM 2024Set eng-2024-06054 marksMCQ
$(x-3)(x+1)+(y-1)(y+3)=0$
›Reveal solutionSolution

Expand and read off centre $\left(-\tfrac g1,-\tfrac f1\right)=(1,-1)$ .

Expand:

$x^2-2x-3+y^2+2y-3=0\;\Rightarrow\;x^2+y^2-2x+2y-6=0.$ …
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KEAM 2024Set eng-2024-06064 marksMCQ
$y$
›Reveal solutionSolution

Equate distances from $(0,k)$ to both points.

Centre is $(0,k)$ . Equal distances to $(-4,3)$ and $(3,-4)$ :
$16+(k-3)^2=9+(k+4)^2.$
$16+k^2-6k+9=9+k^2+8k+16\Rightarrow 25-6k=25+8k\Rightarrow k=0.$ …
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KEAM 2024Set eng-2024-06074 marksMCQ
$\alpha$
›Reveal solutionSolution

For $x^2+y^2+\alpha x-8y+56=0$ , radius $=\sqrt{(\alpha/2)^2+(-4)^2-56}$ . Setting this to $3$ : $(\alpha/2)^2-40=9\Rightarrow(\alpha/2)^2=49\Rightarrow\alpha=\pm14$ .

For $x^2+y^2+2gx+2fy+c=0$ , here $2g=\alpha$ , $2f=-8$ (so $f=-4$ ), $c=56$ . The radius is

$r=\sqrt{g^2+f^2-c}=\sqrt{\left(\tfrac{\alpha}{2}\right)^2+16-56}.$ …
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KEAM 2024Set eng-2024-06084 marksMCQ
$x^2+y^2-2x-4y-4=0$
›Reveal solutionSolution

Comparing with $x^2+y^2+2gx+2fy+c=0$ gives $g=-1,\,f=-2,\,c=-4$ , so radius $=\sqrt{1+4+4}=3$ .

For a circle $x^2+y^2+2gx+2fy+c=0$ , the radius is $\sqrt{g^2+f^2-c}$ . Here

$2g=-2\Rightarrow g=-1,\quad 2f=-4\Rightarrow f=-2,\quad c=-4.$ …
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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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