If one end of a diameter of the circle $x^2 + y^2 - 4x - 6y + 11 = 0$ is $(3, 4)$, then find the coordinate of the other end of the diameter.

If one end of a diameter of the circle x² + y² - 4 – Class 11 Mathematics

NCERT Exemplar · Q7

x^2 + y^2 - 4x - 6y + 11 = 0

Ladakh CbseShort· 2mImportance★★★★★est

65% · 96/148 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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:

$(x + 1)^2 + (y - 4)^2 = 9$ …

Start your 14-day free trial to unlock the full solution →

The center of a circle is the midpoint of any of its diameters. By finding the center from the given equation and using the midpoint formula with the known endpoint, we determine the other endpoint of the diameter to be $\boxed{(1, 2)}$ .

To find the other end of a diameter, we need to understand a fundamental property of circles: the center of the circle is always the midpoint of any diameter. This means if we know the coordinates of the center and one endpoint of a diameter, we can use the midpoint formula to find the coordinates of the other endpoint.

The first step is to extract the center coordinates from the given equation of the circle. The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$ , and its center is at $(-g, -f)$ .

Let's proceed with the steps:

Find the center of the circle.

The given equation of the circle is $x^2 + y^2 - 4x - 6y + 11 = 0$ .

We compare this to the general form $x^2 + y^2 + 2gx + 2fy + c = 0$ .

By comparing the coefficients:

$2g = -4 \implies g = -2$

$2f = -6 \implies f = -3$

The center of the circle, $(h, k)$ , is given by $(-g, -f)$ .

Therefore, the center of the circle is $( -(-2), -(-3) ) = (2, 3)$ .

Tip

Alternatively, you can complete the square to find the standard form $(x-h)^2 + (y-k)^2 = r^2$ .

$x^2 - 4x + y^2 - 6y + 11 = 0$

$(x^2 - 4x + 4) + (y^2 - 6y + 9) + 11 - 4 - 9 = 0$

$(x-2)^2 + (y-3)^2 - 2 = 0$

$(x-2)^2 + (y-3)^2 = 2$

From this standard form, the center is clearly $(2, 3)$ .
Apply the midpoint formula.

Let the known end of the diameter be $A = (x_1, y_1) = (3, 4)$ .

Let the unknown other end of the diameter be $B = (x_2, y_2)$ .

We found the center of the circle to be $C = (h, k) = (2, 3)$ . …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ6 — Find the equation of a circle which touches both the axes and the line…→Next questionQ8 — Find the equation of the circle having (1, -2) as its centre and passi…

Q.If one end of a diameter of the circle x2+y2−4x−6y+11=0x^2 + y^2 - 4x - 6y + 11 = 0x2+y2−4x−6y+11=0 is (3,4)(3, 4)(3,4), then find the coordinate of the other end of the diameter.

🎓Unlock everything free for 14 days

Q.If one end of a diameter of the circle $x^2 + y^2 - 4x - 6y + 11 = 0$ is $(3, 4)$ , then find the coordinate of the other end of the diameter.

Unlock everything free for 14 days