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.
The centre lies on the x-axis, so it is (h,0). Using the distance from centre to (2,3) equals the radius 5, we solve for h and get two circles: (x−6)2+y2=25 and (x+2)2+y2=25.
The standard form of a circle’s equation is (x−h)2+(y−k)2=r2, where (h,k) is the centre and r is the radius. Here, the centre lies on the x-axis, so its y-coordinate is 0. That means the centre is (h,0) for some unknown h. The radius is given as 5, so r2=25.
The circle also passes through (2,3). That point must satisfy the circle’s equation. So we plug it in and solve for h.
Write the general equation
With centre (h,0) and r=5, the equation is
(x−h)2+(y−0)2=25⇒(x−h)2+y2=25.
Substitute the given point
Since (2,3) lies on the circle,
(2−h)2+32=25.
This simplifies to
(2−h)2+9=25⇒(2−h)2=16.
Solve for h
Taking square roots gives
2−h=±4.
So two possibilities:
If 2−h=4, then h=−2.
If 2−h=−4, then h=6.
Both are valid — the centre can be to the left or right of the given point, as long as the distance is 5. …