Q.Find the equation of the circle with centre (−3,2) and radius 4.
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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: Standard form of a circle equation
A circle with centre (h,k) and radius r has the equation (x−h)2+(y−k)2=r2.
Here the centre is (−3,2), so h=−3 and k=2. The radius is r=4.
Substituting into the standard form:
(x−(−3))2+(y−2)2=42
(x+3)2+(y−2)2=16
✓Final answer
The equation of the circle is (x+3)2+(y−2)2=16.
A circle is the set of all points at a fixed distance (radius) from a center; substituting center (−3,2) and radius 4 into the standard form (x−h)2+(y−k)2=r2 gives (x+3)2+(y−2)2=16.
Why the standard form works
The equation of a circle comes directly from the distance formula. If a point (x,y) lies on a circle with center (h,k) and radius r, then its distance from the center must equal r. The distance formula tells us:
(x−h)2+(y−k)2=r
Squaring both sides removes the square root and gives the standard form:
(x−h)2+(y−k)2=r2
This is the fundamental equation of a circle. Every point (x,y) satisfying this equation is exactly r units away from (h,k).
Finding our circle's equation
We have center (h,k)=(−3,2) and radius r=4.
Identify the center coordinates. Here h=−3 and k=2.
Calculate r2. Since r=4, we have r2=16.
Substitute into the standard form. Replace h with −3, k with 2, and r2 with 16:
(x−(−3))2+(y−2)2=16
Simplify the double negative. The term x−(−3) becomes x+3:
(x+3)2+(y−2)2=16
Watch out
Watch the signs carefully. The standard form has (x−h), so when h=−3, you get x−(−3)=x+3, not(x−3)2. A common mistake is writing the center's coordinates with the wrong sign.
The equation is complete. You could expand it to general form x2+y2+6x−4y−3=0 by multiplying out the squares, but the standard form is cleaner and immediately reveals the circle's center and radius.
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
Showing the 12 most recent of 19 on this concept.
KEAM 2026Set eng-2026-04174 marksMCQ
Q.The area of the circle x2+y2+8x−6y+c=0 is 75π . Then the value of c is equal to
(A) -50
(B) 50
(C) 25
(D) -25
(E) -40
›Reveal solutionSolution
For x2+y2+8x−6y+c=0, r2=g2+f2−c; set πr2=75π.
Here 2g=8,2f=−6, so g=4,f=−3 and r2=g2+f2−c=16+9−c=25−c.
The area is πr2=75π, so r2=75. Thus 25−c=75⇒c=−50.
✓Final answer
The correct option is (A).
KEAM 2026Set eng-2026-04194 marksMCQ
Q.If the one end of a diameter of the circle x2+y2+3x+y−6=0 is at (−4,−2), then the other end of the diameter is at
(A) (4,−2)
(B) (1,−1)
(C) (1,1)
(D) (−1,−1)
(E) (1,−2)
›Reveal solutionSolution
The centre bisects any diameter, so the other endpoint is 2C−P1.
For x2+y2+3x+y−6=0, the centre is C=(−23,−21).
With one end P1=(−4,−2), the other end is
P2=2C−P1=(−3−(−4),−1−(−2))=(1,1).
✓Final answer
The correct option is (C).
KEAM 2026Set eng-2026-04204 marksMCQ
Q.The equation of a chord to the circle x2+y2=25 is x+y−5=0. The equation of a circle whose diameter is x+y−5=0, is
(A) x2+y2−6x−5y=0
(B) x2+y2−5x−6y=0
(C) x2+y2−5x−5y=0
(D) x2+y2−6x−6y=0
(E) x2+y2−3x−3y=0
›Reveal solutionSolution
The endpoints of the diameter are the intersections of x+y=5 with x2+y2=25, namely (0,5) and (5,0). The circle with this diameter is (x−0)(x−5)+(y−5)(y−0)=0, i.e. x2+y2−5x−5y=0.
Find where the chord x+y−5=0 meets the circle x2+y2=25. Put y=5−x:
x2+(5−x)2=25⇒2x2−10x=0⇒2x(x−5)=0,
so x=0 (then y=5) or x=5 (then y=0). The endpoints are (0,5) and (5,0).
The circle having a segment as diameter with endpoints (x1,y1),(x2,y2) is
(x−x1)(x−x2)+(y−y1)(y−y2)=0.
Here:
(x−0)(x−5)+(y−5)(y−0)=0⇒x2−5x+y2−5y=0,
i.e. x2+y2−5x−5y=0.
✓Final answer
The correct option is (C).
KEAM 2026Set eng-2026-04214 marksMCQ
Q.The centre and radius of the circle x2+y2−2x+4y=8 respectively are
(A) (1,2),13
(B) (−1,2),13
(C) (−1,−1),13
(D) (1,−2),13
(E) (2,1),13
›Reveal solutionSolution
Completing the square gives centre (1,−2) and radius 13.
For x2+y2−2gx−2fy+c=0 the centre is (g,f) and radius g2+f2−c. Here x2+y2−2x+4y−8=0 gives g=1,f=−2,c=−8:
centre=(1,−2),r=1+4+8=13.
✓Final answer
The correct option is (D).
KEAM 2025Set eng-2025-04234 marksMCQ
Q.A circle touches the x-axis at (9,0). If it also touches the straight line y=14, then the equation of the circle is
(A) (x−9)2+(y−7)2=49
(B) x2+(y−7)2=49
(C) (x−9)2+y2=49
(D) (x−9)2+(y−7)2=81
(E) (x−7)2+(y−9)2=49
›Reveal solutionSolution
The circle is (x−9)2+(y−7)2=49.
Concept and Intuition
A circle tangent to the x-axis at (9,0) has its centre vertically above that point at (9,r). Tangency to a horizontal line above fixes r.
Step-by-Step Solution
Centre =(9,r), radius r (touches x-axis at (9,0)).
Touches y=14: distance ∣14−r∣=r⇒14−r=r⇒r=7.
Centre (9,7), r=7: (x−9)2+(y−7)2=49.
Common Mistakes
Placing the centre off x=9.
Taking r=14 instead of solving 14−r=r.
✓Final answer
The correct option is (A) — (x−9)2+(y−7)2=49.
ANSWER: A
KEAM 2025Set eng-2025-04254 marksMCQ
Q.The equation of the line passing through the point (−4,2) and the centre of the circle 2x2+2y2−8y=7 is
(A) x+3y=2
(B) y=2
(C) x=−4
(D) x+y=−2
(E) y=−2
›Reveal solutionSolution
Find the circle's centre, then the line through it and the given point; both share y=2, so the line is y=2.
Divide the circle equation by 2: x2+y2−4y=27. Complete the square: x2+(y−2)2=27+4. The centre is (0,2).
The required line passes through (−4,2) and the centre (0,2). Both points have y-coordinate 2, so the line is horizontal:
y=2.
✓Final answer
The correct option is (B).
KEAM 2025Set eng-2025-04274 marksMCQ
Q.The centre of the ellipse 4x2+24x+9y2−18y+9=0 is
(A) (1,3)
(B) (1,−3)
(C) (3,−1)
(D) (−3,1)
(E) (3,−3)
›Reveal solutionSolution
Complete the square in x and y; the ellipse becomes 4(x+3)2+9(y−1)2=36, centre (−3,1).
Start from 4x2+24x+9y2−18y+9=0:
4(x2+6x)+9(y2−2y)+9=0.
4(x+3)2−36+9(y−1)2−9+9=0.
4(x+3)2+9(y−1)2=36.
The centre is (−3,1).
✓Final answer
The correct option is (D).
KEAM 2025Set eng-2025-04294 marksMCQ
Q.A circle passes through (4,0) and (0,2) with centre on the y-axis. The radius of the circle is
(A) 5
(B) 10
(C) 15
(D) 20
(E) 25
›Reveal solutionSolution
Centre (0,k) equidistant from (4,0) and (0,2): 16+k2=(k−2)2⇒k=−3; radius =16+9=5.