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 equation is already in standard form (x−h)2+(y−k)2=r2, so we read the centre directly as (−5,3) and the radius as 6.
The standard form of a circle’s equation is the most direct way to find its centre and radius. It’s built from the distance formula: every point (x,y) on the circle is exactly r units away from the centre (h,k). Squaring that distance gives:
(x−h)2+(y−k)2=r2
Here h and k are the x and y coordinates of the centre, and r is the radius.
The given equation is (x+5)2+(y−3)2=36. Notice the signs: x+5 means x−(−5), so h=−5. Similarly, y−3 means y−3, so k=3. The right-hand side is 36, which is r2. So r=36=6.
Identify h from the x-term.
The term (x+5)2 matches (x−h)2 only if x+5=x−(−5). Hence h=−5.
Q.If the one end of a diameter of the circle x2+y2+3x+y−6=0 is at (−4,−2), then the other end of the diameter is at
(A) (4,−2)
(B) (1,−1)
(C) (1,1)
(D) (−1,−1)
(E) (1,−2)
›Reveal solutionSolution
The centre bisects any diameter, so the other endpoint is 2C−P1.
Q.The equation of a chord to the circle x2+y2=25 is x+y−5=0. The equation of a circle whose diameter is x+y−5=0, is
(A) x2+y2−6x−5y=0
(B) x2+y2−5x−6y=0
(C) x2+y2−5x−5y=0
(D) x2+y2−6x−6y=0
(E) x2+y2−3x−3y=0
›Reveal solutionSolution
The endpoints of the diameter are the intersections of x+y=5 with x2+y2=25, namely (0,5) and (5,0). The circle with this diameter is (x−0)(x−5)+(y−5)(y−0)=0, i.e. x2+y2−5x−5y=0.
Find where the chord x+y−5=0 meets the circle x2+y2=25. Put y=5−x:
x2+(5−x)2=25⇒2x2−10x=0⇒2x(x−5)=0,
so x=0 (then y=5) or x=5 (then y=0). The endpoints are (0,5) and (5,0). …
Q.A circle touches the x-axis at (9,0). If it also touches the straight line y=14, then the equation of the circle is
(A) (x−9)2+(y−7)2=49
(B) x2+(y−7)2=49
(C) (x−9)2+y2=49
(D) (x−9)2+(y−7)2=81
(E) (x−7)2+(y−9)2=49
›Reveal solutionSolution
The circle is (x−9)2+(y−7)2=49.
Concept and Intuition
A circle tangent to the x-axis at (9,0) has its centre vertically above that point at (9,r). Tangency to a horizontal line above fixes r.
Step-by-Step Solution
Centre =(9,r), radius r (touches x-axis at (9,0)).
Q.The equation of the line passing through the point (−4,2) and the centre of the circle 2x2+2y2−8y=7 is
(A) x+3y=2
(B) y=2
(C) x=−4
(D) x+y=−2
(E) y=−2
›Reveal solutionSolution
Find the circle's centre, then the line through it and the given point; both share y=2, so the line is y=2.
Divide the circle equation by 2: x2+y2−4y=27. Complete the square: x2+(y−2)2=27+4. The centre is (0,2). …