Q.Equation of the circle with centre on the y-axis and passing through the origin and the point (2,3) is
(A) x2+y2+13y=0
(B) 3x2+3y2+13x+3=0
(C) 6x2+6y2−13x=0
(D) x2+y2+13x+3=0
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:
(x−h)2+(y−k)2=r
Now square both sides to remove the square root:
(x−h)2+(y−k)2=r2
That's it. That's the standard form of the equation of a circle.
(x−h)2+(y−k)2=r2
where (h,k) is the centre and r is the radius (r>0).
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 x2+y2=r2.
r2 — notice it's the square of the radius, not the radius itself. If the equation says x2+y2=25, the radius is 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.
Taking the centre on the axis stated in the stem, through the origin and (2,3), the circle is 3x2+3y2−13y=0. (The original NCERT Exemplar places the centre on the x-axis, giving 2x2+2y2−13x=0, the intended option (C); the printed options here are corrupted.)
A general circle passing through the origin has no constant term:
x2+y2+2gx+2fy=0,
with centre (−g,−f).
Stem as printed — centre on the y-axis.
A centre on the y-axis means the x-coordinate of the centre is 0, i.e. −g=0, so g=0 and the equation reduces to x2+y2+2fy=0. It passes through (2,3):
22+32+2f(3)=0⇒13+6f=0⇒f=−613.
Hence x2+y2−313y=0; multiplying through by 3,
3x2+3y2−13y=0.
Check: origin →0; (2,3)→3(4)+3(9)−13(3)=12+27−39=0. Both points lie on it. …