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.
Concept: Circle Equation Standard Form and External Tangency Condition
The given circle x2+y2−2x−4y−20=0 can be rewritten as (x−1)2+(y−2)2=25, so its center is C1=(1,2) with radius r1=5.
Since the required circle (center C2, radius r2=5) touches the given circle at (5,5), the point of tangency lies on the line joining the two centers. For external tangency, ∣C1C2∣=r1+r2=10.
The center C2 lies on the line through C1(1,2) and (5,5). The direction vector is (5−1,5−2)=(4,3) with magnitude 5. Extending this by distance 10 from C1: …
The given circle has centre (1,2) and radius 5; the point (5,5) lies on it. The required circle (radius 5) touches at (5,5) with its centre on the line of centres, at (9,8), giving (x−9)2+(y−8)2=25.
When two circles touch, the point of contact and the two centres are collinear (they share a common tangent there).
1. Put the given circle in standard form.
x2+y2−2x−4y−20=0⇒(x−1)2+(y−2)2=25
So its centre is C1=(1,2) and radius r1=5.
2. Confirm the point of contact lies on it.
(5−1)2+(5−2)2=16+9=25
3. Direction of the line of centres.
C1P=(5−1,5−2)=(4,3),∣C1P∣=5,u^=(54,53)
4. Locate the new centre C2 (radius 5, at distance 5 from P).
Moving from P=(5,5) away from C1 (external contact, so ∣C1C2∣=r1+r2=10): …