The line $x + 3y = 0$ is a diameter of the circle $x^2 + y^2 + 6x + 2y = 0$.

The line x + 3y = 0 is a diameter of the circle x² + y² + 6x + 2y = 0.

NCERT Exemplar · Q52

x + 3y = 0

CBSEShort· 1mImportance★★★★★est

95% · 141/148 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 →

A diameter of a circle must pass through its center. By finding the center of the given circle and substituting its coordinates into the line equation, we find that the center does not lie on the line, meaning the statement is false.

To determine if a given line is a diameter of a circle, we need to understand the fundamental property of a diameter: it is any straight line segment that passes through the center of the circle and whose endpoints lie on the circle. This means that if a line is a diameter, it must contain the center of the circle.

Our approach will be to first find the center of the given circle and then check if this center point satisfies the equation of the given line. If it does, the line passes through the center and thus could be a diameter. If it doesn't, the line cannot be a diameter.

Identify the standard form of a circle equation and extract the center.

The general equation of a circle is given by $x^2 + y^2 + 2gx + 2fy + c = 0$ .

For a circle in this form, its center is at the point $(-g, -f)$ .

The given circle equation is $x^2 + y^2 + 6x + 2y = 0$ .

Comparing this to the general form:

$2g = 6 \implies g = 3$

$2f = 2 \implies f = 1$

The constant term $c$ is $0$ in this case, but it doesn't affect the center coordinates.

Therefore, the center of the circle is $(-g, -f) = (-3, -1)$ .
$x^2 + y^2 + 2gx + 2fy + c = 0$
Check if the circle's center lies on the given line.

The given line is $x + 3y = 0$ .

For the line to be a diameter, the center of the circle, which is $(-3, -1)$ , must satisfy the equation of the line.

Substitute $x = -3$ and $y = -1$ into the line equation: …

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

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 21 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$x^2 + y^2 + 8x + 10y - 8 = 0$
(a) $(8, 10)$
(b) $(4, 5)$
(c) $(-4, 5)$
(d) $(-4, -5)$
›Reveal solutionSolution

Match the given equation to the general circle form $x^2+y^2+2gx+2fy+c=0$ and read off the centre $(-g,-f)$ .

The general equation of a circle is $x^2+y^2+2gx+2fy+c=0$ , with centre $(-g,-f)$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$x^2 + y^2 = a^2$
(a) $(1, 1)$
(b) $(1, 0)$
(c) $(0, 0)$
(d) $(0, 1)$
›Reveal solutionSolution

The centre is $(0,0)$ , option (c).

The general equation of a circle with centre $(h,k)$ and radius $r$ is $(x-h)^2+(y-k)^2=r^2$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. The centre of the circle given by the equation x² + y² - 8x + 12y - 12 = 0 has the coordinates:$
(a) (8, -12)
(b) (-8, 12)
(c) (4, -6)
(d) (-4, 6)
›Reveal solutionSolution

Comparing with the general circle equation x²+y²+2gx+2fy+c=0, the centre is (−g, −f).

The general form of a circle is:

$x^2+y^2+2gx+2fy+c=0, \quad \text{centre} = (-g,-f)$

Given $x^2+y^2-8x+12y-12=0$ , compare coefficients: …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$x^2 + y^2 = 256$
(a) 16
(b) 32
(c) 8
(d) 64
›Reveal solutionSolution

Diameter of $x^2+y^2=256$ is $32$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$x^2 + y^2 = 9$
(a) $(9, 0)$
(b) $(0, 9)$
(c) $(9, 9)$
(d) $(0, 0)$
›Reveal solutionSolution

Centre of $x^2+y^2=9$ is $(0,0)$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$A(2, -9)$
(a) $(2, -4)$
(b) $(-3, 4)$
(c) $\left(3, \dfrac{-16}{3}\right)$
(d) None of these
›Reveal solutionSolution

Substituting all three points into $x^2+y^2+2gx+2fy+c=0$ and solving the resulting linear system gives centre $(-g,-f) = (2,-4)$ .

Let the circle be $x^2+y^2+2gx+2fy+c=0$ .

Substitute $A(2,-9)$ : $4+81+4g-18f+c=0 \implies 4g-18f+c=-85$ ...(1)

Substitute $B(5,-8)$ : $25+64+10g-16f+c=0 \implies 10g-16f+c=-89$ ...(2)

Substitute $C(2,1)$ : $4+1+4g+2f+c=0 \implies 4g+2f+c=-5$ ...(3)

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$x^2+y^2-8x+10y-2=0$
(a) $(4,-5)$
(b) $(-5,4)$
(c) $(5,3)$
(d) None of these
›Reveal solutionSolution

Compare the given equation to the general circle form $x^2+y^2+Dx+Ey+F=0$ , whose centre is $\left(-\dfrac D2, -\dfrac E2\right)$ .

The general equation of a circle is:

$x^2 + y^2 + Dx + Ey + F = 0, \quad \text{centre} = \left(-\frac{D}{2}, -\frac{E}{2}\right)$

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$(2,2)$
(a) $x^2+y^2+4x+4y-5=0$
(b) $x^2+y^2-4x-4y-5=0$
(c) $x^2+y^2=27$
(d) None of these
›Reveal solutionSolution

Find the radius using the distance from centre to the given point, then write $(x-h)^2+(y-k)^2=r^2$ and expand.

The centre is $(2,2)$ and the circle passes through $(4,5)$ , so the radius is the distance between them:

$r^2 = (4-2)^2 + (5-2)^2 = 2^2 + 3^2 = 4 + 9 = 13$

The circle's equation is:

$(x-2)^2 + (y-2)^2 = 13$

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 mark
$(-3, 2)$
›Reveal solutionSolution

The required circle's equation is $(x+3)^2+(y-2)^2=16$ .

The standard equation of a circle with centre $(h,k)$ and radius $r$ is $(x-h)^2+(y-k)^2=r^2$ . Substituting the centre $(-3,2)$ and radius $4$ : …

Start your 14-day free trial to see the full solution →
CBSE 2024Set sz1 markMCQ
$(0, 2)$
(a) $x^2 + y^2 - 6y = 0$
(b) $x^2 + y^2 - 4y = 0$
(c) $x^2 + 2y^2 - 3 = 0$
(d) $2x^2 + y^2 + 3y = 0$
›Reveal solutionSolution

The circle with centre $(0,2)$ and radius $2$ expands to $x^2+y^2-4y=0$ .

The standard equation of a circle with centre $(h,k)$ and radius $r$ is:

$(x-h)^2 + (y-k)^2 = r^2$

Here $h=0, k=2, r=2$ :

$(x-0)^2 + (y-2)^2 = 2^2$ …

Start your 14-day free trial to see the full solution →
CBSE 2023Set ANNUAL1 markMCQ
$(0, 1)$
(a) $x^2+y^2=16$
(b) $x^2+(y-1)^2=16$
(c) $(x-1)^2+y^2=16$
(d) $x^2+(y+1)^2=16$
›Reveal solutionSolution

Plugging centre $(0,1)$ and radius $4$ into the standard circle equation gives $x^2+(y-1)^2=16$ .

The standard equation of a circle with centre $(h,k)$ and radius $r$ is:

$(x-h)^2+(y-k)^2=r^2$

…

Start your 14-day free trial to see the full solution →
CBSE 2023Set ANNUAL1 markMCQ
$(1,1)$
(a) $x^2+y^2-2x-2y=0$
(b) $x^2+y^2+2x+2y=0$
(c) $(x+1)^2+(y+1)^2=2$
(d) none of these
›Reveal solutionSolution

The circle with centre $(1,1)$ and radius $\sqrt2$ has equation $x^2+y^2-2x-2y=0$ .

The standard equation of a circle with centre $(h,k)$ and radius $r$ is $(x-h)^2+(y-k)^2=r^2$ . Here $h=1,k=1,r=\sqrt2$ so $r^2=2$ :

$(x-1)^2+(y-1)^2=2$

Expanding:

…

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

←Previous questionQ51 — The equation of the hyperbola with vertices at (0, ± 6) and eccentrici…→Next questionQ53 — The shortest distance from the point (2, -7) to the circle x² + y² - 1…

Q.The line x+3y=0x + 3y = 0x+3y=0 is a diameter of the circle x2+y2+6x+2y=0x^2 + y^2 + 6x + 2y = 0x2+y2+6x+2y=0.

🎓Unlock everything free for 14 days

Q.The line $x + 3y = 0$ is a diameter of the circle $x^2 + y^2 + 6x + 2y = 0$ .

Unlock everything free for 14 days