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.
Here h=−1, k=4, r=3. Plug in:
(x−(−1))2+(y−4)2=32
Simplify:
(x+1)2+(y−4)2=9
That's the standard form. From this, you can immediately read off the centre (−1,4) and radius 3.
Why This Form Matters
The standard form is the most useful because it gives you the centre and radius at a glance. In exams, you'll often be given an expanded form like x2+y2−6x+4y−12=0 and asked to rewrite it in standard form by completing the square — that's the next step in your learning, but the standard form itself is the destination.
For now: centre tells you where, radius tells you how big, and the equation tells you which points belong.
The Standard Form of a Circle's Equation is one of the first results in the NCERT Class 11 Mathematics chapter on Conic Sections, matching searches like "equation of a circle: definition, formula and examples" or "conic sections important questions class 11 maths". Recognising centre and radius directly from this form is also a routine, quick-scoring question type in CBSE boards, JEE Main, and state CET coordinate geometry sections.
Concept: Circle Equation Standard Form
The standard form is (x−h)2+(y−k)2=r2, where (h,k) is the centre and r is the radius.
Steps
Identify h=−2, k=3, r=4.
Substitute into the formula:
(x−(−2))2+(y−3)2=42
Simplify:
(x+2)2+(y−3)2=16
✓Final answer
The equation is (x+2)2+(y−3)2=16.
The equation of a circle is derived from its geometric definition as the set of points at a fixed distance (radius) from a fixed point (centre). Using the distance formula, we get (x+2)2+(y−3)2=16.
The standard form of a circle’s equation comes straight from the definition of a circle: every point (x,y) on the circle is exactly r units away from the centre (h,k). That distance is given by the Euclidean distance formula. So instead of memorising a formula, think of it as “distance from centre equals radius” — that’s all there is.
Write the distance condition.
Let the centre be C(−2,3) and radius r=4. For any point P(x,y) on the circle, the distance CP must equal 4. Using the distance formula:
(x−(−2))2+(y−3)2=4
which simplifies to:
(x+2)2+(y−3)2=4
Remove the square root.
Square both sides to eliminate the radical. This is valid because both sides are non-negative:
(x+2)2+(y−3)2=16
Interpret the result.
This is the equation of the circle in standard form (x−h)2+(y−k)2=r2, where h=−2, k=3, and r2=16. Notice the signs: (x+2) is (x−(−2)), so the centre is indeed (−2,3).
Watch out
A common mistake is to write (x−2)2+(y+3)2=16 by forgetting that x−(−2) becomes x+2, not x−2. Always check the sign of the centre coordinates.
Tip
If you ever forget the form, just re-derive it from the distance definition — it takes 10 seconds and is foolproof.
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
COMEDK 2025Set 2025-M1 markMCQ
Q.Equation of a circle whose area is 154 sq units and having 2x−3y+12=0 and x+4y−5=0 as diameters is
(A) x2+y2+6x−4y+36=0
(B) x2−y2+6x−4y−36=0
(C) x2+y2−6x+4y−36=0
(D) x2+y2+6x−4y−36=0
›Reveal solutionSolution
The circle's center is the intersection of its two given diameters, and its radius comes from the given area. This gives x2+y2+6x−4y−36=0, matching option (D).
Step-by-step reasoning
Find the centre (intersection of the diameters).
2x−3y+12=0,x+4y−5=0
From the second: x=5−4y. Substituting into the first:
2(5−4y)−3y+12=0⇒10−8y−3y+12=0⇒22−11y=0⇒y=2
Then x=5−4(2)=−3. Centre =(−3,2).
Find the radius from the area.
πr2=154(π≈722)⇒r2=154×227=49⇒r=7
Write and expand the circle equation.
(x+3)2+(y−2)2=49
x2+6x+9+y2−4y+4=49⇒x2+y2+6x−4y−36=0
Watch out
The centre is the intersection of the two diameters, not just any point on either line.
Tip
Verify by substituting the centre into the final equation: it should give −r2. At (−3,2): 9+4−18−8−36=−49=−r2✓.
✓Final answer
The correct option is (D).
KCET 2021Set A-11 markMCQ
Q.If the parabola y=αx2−6x+β passes through the point (0,2) and has its tangent at x=23 parallel to x axis, then
(A) α=2,β=−2
(B) α=−2,β=2
(C) α=2,β=2
(D) α=−2,β=−2
›Reveal solutionSolution
The parabola passes through (0,2) and has a horizontal tangent at x=23. Using the point condition gives β=2, and the derivative condition gives α=2. The correct pair is α=2,β=2, option (C).
The key idea here is that a tangent parallel to the x-axis means the slope of the tangent is zero. For a curve y=f(x), the slope of the tangent at any point is given by dxdy. So "tangent parallel to x-axis" translates directly to dxdy=0 at that x-coordinate.
We also have a point that lies on the parabola. Substituting that point into the equation gives a direct relation between α and β.
Let’s work through it step by step.
Use the point (0,2).
The parabola is y=αx2−6x+β. Substituting x=0, y=2:
2=α(0)2−6(0)+β⇒β=2.
Find the derivative.
Differentiate y with respect to x:
dxdy=2αx−6.
Apply the tangent condition.
At x=23, the tangent is parallel to the x-axis, so dxdy=0:
2α(23)−6=0⇒3α−6=0⇒α=2.
Check the options.
We have α=2, β=2. This matches option (C).
Watch out
A common mistake is to forget that "parallel to x-axis" means slope zero, not just "horizontal" in the sense of a constant function. Also, be careful: the derivative is set to zero at the given x, not at the given point (0,2).
Tip
You could also think of this geometrically: a parabola y=ax2+bx+c has its vertex where the tangent is horizontal. Here, the vertex occurs at x=23, so you could directly use the vertex formula x=−2ab with b=−6 to get −2α−6=23, which gives α=2 — same result, faster.
✓Final answer
The correct option is (C), with α=2 and β=2.
COMEDK 2021Set 20211 markMCQ
Q.What will be the equation of circle whose centre is (1, 2) and touches X-axis?
(A) x2+y2−2x−4y+1=0
(B) x2−y2+2x+4y+1=0
(C) x2+y2+2x−4y−1=0
(D) x2+y2+2x+4y−1=0
›Reveal solutionSolution
Check: centre = (1, 2), r = sqrt(1 + 4 - 1) = sqrt(4) = 2. Distance from centre to the X-axis = 2 = r, so it touches the X-axis. Correct.
Concept: A circle that touches the X-axis has radius equal to the absolute value of the y-coordinate of its centre (the perpendicular distance from the centre to the line y = 0).
Centre (h, k) = (1, 2) -> radius r = |k| = 2.
Equation: (x - 1)^2 + (y - 2)^2 = 2^2
x^2 - 2x + 1 + y^2 - 4y + 4 = 4
x^2 + y^2 - 2x - 4y + 1 = 0
Check: centre = (1, 2), r = sqrt(1 + 4 - 1) = sqrt(4) = 2. Distance from centre to the X-axis = 2 = r, so it touches the X-axis. Correct.
✓Final answer
The correct option is (A) — x2+y2−2x−4y+1=0
ANSWER: A
COMEDK 2021Set 20211 markMCQ
Q.Find the centre and radius of the circle given by the equation 2x2+2y2+3x+4y+89=0.
(A) 1
(B) −1
(C) 2
(D) −2
›Reveal solutionSolution
The four options are single numbers, so they refer to the radius; the radius is 1.
Concept: general equation of a circle x^2 + y^2 + 2gx + 2fy + c = 0 has centre (-g, -f) and radius sqrt(g^2 + f^2 - c).
First make the coefficients of x^2 and y^2 equal to 1 by dividing throughout by 2:
2x^2 + 2y^2 + 3x + 4y + 9/8 = 0
=> x^2 + y^2 + (3/2)x + 2y + 9/16 = 0.
So 2g = 3/2 => g = 3/4; 2f = 2 => f = 1; c = 9/16.
The four options are single numbers, so they refer to the radius; the radius is 1.
✓Final answer
The correct option is (A) — 1
ANSWER: A
COMEDK 2021Set 20211 markMCQ
Q.What will be the equation of the circle whose centre is (1, 2) and which passes through the point (4, 6)?
(A) x2+y2−2x−4y−20=0
(B) x2+y2+2x+4y−20=0
(C) x2−y2−2x−4y+20=0
(D) x2−y2+2x−4y−20=0
Q.The equation of the line parallel to the line 3x−4y+2=0 and passing through (−2,3) is
(A) 3x−4y+18=0
(B) 3x−4y−18=0
(C) 3x+4y+18=0
(D) 3x+4y−18=0
›Reveal solutionSolution
Parallel lines have equal slopes, so keep the x- and y-coefficients unchanged and fix the constant by making the line pass through the given point.
Step 1 — Use the parallelism condition.
The slope of 3x−4y+2=0 is
m=−coefficient of ycoefficient of x=−−43=43
Any line parallel to it has the same slope, so it can be written by changing only the constant term:
3x−4y+c=0
This is the standard "family of parallel lines" trick — it saves converting to slope-intercept form.
Step 2 — Impose the point condition.
The line must pass through (−2,3), so those coordinates must satisfy it:
3(−2)−4(3)+c=0
−6−12+c=0⇒c=18
Step 3 — Write the equation.
3x−4y+18=0
Step 4 — Verify.
Substitute (−2,3): 3(−2)−4(3)+18=−6−12+18=0✓. And its slope is 3/4, matching the given line. ✓
Options (C) and (D) have +4y, giving slope −3/4 — those are lines perpendicular-ish in coefficient pattern, certainly not parallel. Option (B) has the right direction but fails the point test: −6−12−18=−36=0.