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 — (x−h)2+(y−k)2=r2.
Step 1: Identify centre (h,k)=(21,41) and radius r=121.
Step 2: Substitute into the standard form:
(x−21)2+(y−41)2=(121)2
Step 3: Square the radius:
(121)2=1441
✓Final answer
The equation is (x−21)2+(y−41)2=1441.
The standard form of a circle is (x−h)2+(y−k)2=r2. Substituting the given centre (21,41) and radius 121, then simplifying, gives the equation 36x2+36y2−36x−18y+11=0.
The equation of any circle is built from its centre and radius. If you know the centre (h,k) and the radius r, the circle is the set of all points (x,y) that are exactly r units away from (h,k). That distance condition is just the Pythagorean theorem in disguise.
(x−h)2+(y−k)2=r2
This is the standard form of a circle. It’s the most direct way to write the equation when you’re given the centre and radius. No shifting, no completing the square — just plug in and simplify.
Substitute the centre and radius.
Here h=21, k=41, and r=121.
(x−21)2+(y−41)2=(121)2
Square the radius.(121)2=1441. So we have:
(x−21)2+(y−41)2=1441
Expand the squares.
(x−21)2=x2−x+41
(y−41)2=y2−21y+161
Adding them:
x2+y2−x−21y+41+161=1441
Combine the constant terms.41=164, so 41+161=165.
The equation becomes:
x2+y2−x−21y+165=1441
Move the constant to the right side.
x2+y2−x−21y=1441−165
Compute the right side. 165=14445, so:
1441−14445=−14444=−3611
Thus:
x2+y2−x−21y=−3611
Clear the fractions by multiplying through by 36.
36x2+36y2−36x−18y=−11
Bring everything to one side.
36x2+36y2−36x−18y+11=0
Watch out
A common mistake is forgetting to square the radius, or mishandling the fractions when combining 41 and 161. Always write every term with a common denominator before adding or subtracting — it saves errors.
Tip
If you prefer to avoid fractions entirely, multiply the standard form by the least common multiple of the denominators (here 144) right after substitution. That gives integer coefficients from the start, though the expansion is a bit heavier. The method above keeps the algebra cleaner step by step.
✓Final answer
The equation of the circle is 36x2+36y2−36x−18y+11=0.
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.