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.
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 …
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). …
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).
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