Q.Find the equation of circle passing through (3,4), (3,2), (1,4).
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🔒 Start your 14-day free trial to unlock the full solution →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.
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 …
Substitute the three points into x2+y2+2gx+2fy+c=0 and solve the resulting linear system for g,f,c. …
Solving the three conditions gives g=−4,f=−3,c=23, so the circle is x2+y2−8x−6y+23=0.
Let the circle be x2+y2+2gx+2fy+c=0. Substituting each point:
(3,4):25+6g+8f+c=0⇒6g+8f+c=−25(1)
(3,2):13+6g+4f+c=0⇒6g+4f+c=−13(2)
(1,4):17+2g+8f+c=0⇒2g+8f+c=−17(3)
(1)−(2):4f=−12⇒f=−3.
…
Showing the 12 most recent of 21 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.Centre of the circle x2+y2+8x+10y−8=0 is(a) (8,10)(b) (4,5)(c) (−4,5)(d) (−4,−5)
›Reveal solutionSolution
Match the given equation to the general circle form x2+y2+2gx+2fy+c=0 and read off the centre (−g,−f).
The general equation of a circle is x2+y2+2gx+2fy+c=0, with centre (−g,−f).
…
- CBSE 2026Set ANNUAL1 markMCQQ.The coordinates of the centre of the circle x2+y2=a2 is —(a) (1,1)(b) (1,0)(c) (0,0)(d) (0,1)
›Reveal solutionSolution
The centre is (0,0), option (c).
The general equation of a circle with centre (h,k) and radius r is (x−h)2+(y−k)2=r2.
…
- CBSE 2026Set ANNUAL1 markMCQQ.The centre of the circle given by the equation x² + y² - 8x + 12y - 12 = 0 has the coordinates:(a) (8, -12)(b) (-8, 12)(c) (4, -6)(d) (-4, 6)
›Reveal solutionSolution
Comparing with the general circle equation x²+y²+2gx+2fy+c=0, the centre is (−g, −f).
The general form of a circle is:
x2+y2+2gx+2fy+c=0,centre=(−g,−f)
Given x2+y2−8x+12y−12=0, compare coefficients: …
- CBSE 2025Set ANNUAL1 markMCQQ.The diameter of the circle x2+y2=256 is(a) 16(b) 32(c) 8(d) 64
›Reveal solutionSolution
Diameter of x2+y2=256 is 32.
…
- CBSE 2025Set ANNUAL1 markMCQQ.The centre of the circle x2+y2=9 is(a) (9,0)(b) (0,9)(c) (9,9)(d) (0,0)
›Reveal solutionSolution
Centre of x2+y2=9 is (0,0).
…
- CBSE 2025Set ANNUAL1 markMCQQ.A circle passes through the points A(2,−9), B(5,−8) and C(2,1). The centre of the circle is(a) (2,−4)(b) (−3,4)(c) (3,3−16)(d) None of these
›Reveal solutionSolution
Substituting all three points into x2+y2+2gx+2fy+c=0 and solving the resulting linear system gives centre (−g,−f)=(2,−4).
Let the circle be x2+y2+2gx+2fy+c=0.
Substitute A(2,−9): 4+81+4g−18f+c=0⟹4g−18f+c=−85 ...(1)
Substitute B(5,−8): 25+64+10g−16f+c=0⟹10g−16f+c=−89 ...(2)
Substitute C(2,1): 4+1+4g+2f+c=0⟹4g+2f+c=−5 ...(3)
…
- CBSE 2024Set ANNUAL1 markMCQQ.Find the centre of the circle whose equation is x2+y2−8x+10y−2=0.(a) (4,−5)(b) (−5,4)(c) (5,3)(d) None of these
›Reveal solutionSolution
Compare the given equation to the general circle form x2+y2+Dx+Ey+F=0, whose centre is (−2D,−2E).
The general equation of a circle is:
x2+y2+Dx+Ey+F=0,centre=(−2D,−2E)
…
- CBSE 2024Set ANNUAL1 markMCQQ.Find the equation of the circle whose centre is (2,2) and passes through the point (4,5).(a) x2+y2+4x+4y−5=0(b) x2+y2−4x−4y−5=0(c) x2+y2=27(d) None of these
›Reveal solutionSolution
Find the radius using the distance from centre to the given point, then write (x−h)2+(y−k)2=r2 and expand.
The centre is (2,2) and the circle passes through (4,5), so the radius is the distance between them:
r2=(4−2)2+(5−2)2=22+32=4+9=13
The circle's equation is:
(x−2)2+(y−2)2=13
…
- CBSE 2024Set ANNUAL1 markQ.Find the equation of the circle with centre (−3,2) and radius 4 unit.
›Reveal solutionSolution
The required circle's equation is (x+3)2+(y−2)2=16.
The standard equation of a circle with centre (h,k) and radius r is (x−h)2+(y−k)2=r2. Substituting the centre (−3,2) and radius 4: …
- CBSE 2024Set sz1 markMCQQ.Equation of circle with centre (0,2) and radius 2 is equal to:(a) x2+y2−6y=0(b) x2+y2−4y=0(c) x2+2y2−3=0(d) 2x2+y2+3y=0
›Reveal solutionSolution
The circle with centre (0,2) and radius 2 expands to x2+y2−4y=0.
The standard equation of a circle with centre (h,k) and radius r is:
(x−h)2+(y−k)2=r2
Here h=0,k=2,r=2:
(x−0)2+(y−2)2=22 …
- CBSE 2023Set ANNUAL1 markMCQQ.The equation of the circle whose radius is 4 and centre is (0,1) is(a) x2+y2=16(b) x2+(y−1)2=16(c) (x−1)2+y2=16(d) x2+(y+1)2=16
›Reveal solutionSolution
Plugging centre (0,1) and radius 4 into the standard circle equation gives x2+(y−1)2=16.
The standard equation of a circle with centre (h,k) and radius r is:
(x−h)2+(y−k)2=r2
…
- CBSE 2023Set ANNUAL1 markMCQQ.The equation of the circle with centre (1,1) and radius 2 units will be:(a) x2+y2−2x−2y=0(b) x2+y2+2x+2y=0(c) (x+1)2+(y+1)2=2(d) none of these
›Reveal solutionSolution
The circle with centre (1,1) and radius 2 has equation x2+y2−2x−2y=0.
The standard equation of a circle with centre (h,k) and radius r is (x−h)2+(y−k)2=r2. Here h=1,k=1,r=2 so r2=2:
(x−1)2+(y−1)2=2
Expanding:
…
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