Math101 · Independent Ontario learning library
Math101learn.math101.caCircle Equation
A circle with centre $(h,k)$ and radius $r>0$ has equation $(x-h)^2+(y-k)^2=r^2$. Every solution point is distance $r$ from the centre, so the equation is the distance formula with the square root removed.
Circle equations translate fixed-distance geometry into algebra, enabling coordinate proofs, intersections, loci, and analytic geometry.
Intuition and core definition
A circle with centre $(h,k)$ and radius $r>0$ has equation $(x-h)^2+(y-k)^2=r^2$. Every solution point is distance $r$ from the centre, so the equation is the distance formula with the square root removed.
Notation, language, and conditions
Signs reverse inside grouped coordinates: $(x+3)^2$ corresponds to centre coordinate $h=-3$. General form $x^2+y^2+Dx+Ey+F=0$ can be converted by completing the square. Equal coefficients on $x^2,y^2$ and no $xy$ term characterize an axis-aligned circle equation after scaling.
Why this idea matters
The standard circle equation records squared distance from a centre, with radius determining the common distance of every point on the curve.
A dependable method
- Read centre and radius from standard form, or group $x$ and $y$ terms in general form.
- Complete the square in each variable, adding the same quantities to both sides.
- Write the equation as two squared binomials equal to a positive number.
- Take the positive square root of the right side for radius.
- Substitute centre offsets or a claimed point to check.
Worked example
Representations and interpretation
The equation is a level set of squared distance. A graph shows all directions around the centre simultaneously; horizontal extremes are $(h\pm r,k)$ and vertical extremes $(h,k\pm r)$.
Reasoning about variations
If the completed-square right side is zero, the locus is a single point rather than a nondegenerate circle. If it is negative, there are no real points satisfying the equation.
Common mistakes
How to check your work
- Plot the centre and four axis extremes.
- Substitute one extreme point.
- Expand standard form and recover the general equation.
Practice
- State the centre and radius of $(x+2)^2+(y-5)^2=49$.
- Write the circle with centre $(1,-3)$ and radius $4$.
- Does $(3,4)$ lie on $x^2+y^2=25$?
Answers and brief solutions
Show answers
- Centre $(-2,5)$; radius $7$ The grouped signs are opposite the centre coordinates, and $r=\sqrt{49}$.
- $(x-1)^2+(y+3)^2=16$ Use squared coordinate differences equal to $r^2$.
- Yes $3^2+4^2=25$.
Synthesis and transfer
A coverage boundary around a transmitter can be modelled by a circle equation; substituting a location shows whether it lies on the boundary, inside, or outside.
A transmitter centred at $(4,-1)$ with range $6$ has boundary $(x-4)^2+(y+1)^2=36$. A location makes the left side less than $36$ when it is inside coverage, equal when it is on the boundary, and greater when it is outside. The signs inside the squares are opposite the centre coordinates because the equation measures coordinate differences. Expanding hides the centre but does not change the locus; completing squares restores the geometric information. A range model may use the filled disk rather than only the circle, so the contextual inequality must be distinguished from the boundary equation.
Related topics
Teaching and accessibility note
Try it yourself
Hints are part of learning. Open one whenever it makes the next step feel possible.
State the centre and radius of $(x+2)^2+(y-5)^2=49$.
- The grouped signs are opposite the centre coordinates, and $r=\sqrt{49}$.
End of lesson
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