Circles MCQ Quiz - Objective Question with Answer for Circles - Download Free PDF
Last updated on Jun 14, 2025
Latest Circles MCQ Objective Questions
Circles Question 1:
A square is inscribed in a circle x 2 + y 2 + 2x + 2y + 1 = 0 and its sides are parallel to coordinate axes. Which one of the following is a vertex of the square?
Answer (Detailed Solution Below)
Circles Question 1 Detailed Solution
Calculation:
Given,
The circle’s equation is
Rewrite by completing squares:
∴ center = (-1, -1), radius r = 1
For a square inscribed in this circle with sides parallel to the axes, its diagonal equals the circle’s diameter = 2. If the square has side length a, then
Each vertex lies a half‐side
One of these vertices is
Hence, the correct answer is Option 3
Circles Question 2:
What is the equation of the circle whose diameter is 10 cm and the equations of two of its diameters are x+y=0 and x−y=0?
Answer (Detailed Solution Below)
Circles Question 2 Detailed Solution
Calculation:
Given two diameters of the circle are along the lines x + y = 0 and x - y = 0. The center of the circle is at their intersection.
We have,
Substitute into x + y = 0:
Thus, the center is (0, 0).
Also radius,
the equation of the circle with center (0,0) and radius 5 is
Hence, the correct answer is Option 2.
Circles Question 3:
If the four distinct points (4, 6), (–1, 5), (0, 0) and (k, 3k) lie on a circle of radius r, then 10k + r2 is equal to
Answer (Detailed Solution Below)
Circles Question 3 Detailed Solution
Calculation:
m1m2 = – 1 so right angle equation circle is
⇒ (x – 4) (x – 0) + (y – 6) (y – 0) = 0
⇒ x2 + y2 – 4x – 6y = 0
⇒ (k,3k) lies on it so
⇒ k2 + 9k2 – 4k – 18k = 0
⇒ 10k2 – 22k = 0
⇒ k = 0,
k = 0 is not possible so k =
also r =
so 10k + r2 =
Hence, the Correct answer is Option 4.
Circles Question 4:
The absolute difference between the squares of the radii of the two circles passing through the point (–9, 4) and touching the lines x + y = 3 and x – y = 3, is equal to _____.
Answer (Detailed Solution Below) 1 - 768
Circles Question 4 Detailed Solution
Concept:
Circle Equation and Radius:
- The radius of a circle touching a line can be found using the perpendicular distance from the center to the line.
- The circle equation is
, where is the center and is the radius.
Calculation:
Center of circle is
Radius
Equation of circle:
Circle passes through point
Expanding and simplifying:
Factorizing:
Calculating radii:
Absolute difference between squares of radii:
Hence, the correct answer is 768.
Circles Question 5:
Let C be the circle of minimum area touching the parabola y = 6 – x2 and the lines y = √3 |x| . Then, which one of the following points lies on the circle
Answer (Detailed Solution Below)
Circles Question 5 Detailed Solution
Ans. (1)
Sol.
Equation of circle
x2 + (y – (6 – r))2 = r2
touches
p = r
|r – 6| = 2r
r = 2
∴ Circle x2 + (y – 4)2 = 4
(2, 4) Satisfies this equation
Top Circles MCQ Objective Questions
AB is a chord in the minor segment of a circle with center O. C is a point between A and B on the minor arc AB. The tangents to the circle at A and B meet at the point D. If ∠ACB = 116°, then the measure of ∠ADB is
Answer (Detailed Solution Below)
Circles Question 6 Detailed Solution
Download Solution PDFGiven:
∠ACB = 116°
Concept Used:
When a quadrilateral is inscribed in a circle, the opposite angles of it are supplementary angles.
The angle subtended at the center is always double the angle subtended at the remaining arc.
The radius from the center of the circle to the point of tangency is perpendicular to the tangent line.
Calculation:
Point P is taken on the major arc of the circle.
Then, A & P, B & P, C & B, and A & C are joined.
From cyclic quadrilateral APBC
∠ACB = 116°
Now, ∠APB = (180 – 116)° = 64°
Now, ∠AOB = (64 × 2)°
⇒ 128°
Since OA = OB = radius of the circle
So, from quadrilateral AOBD
∠ADB = 360° - (∠OBD + ∠OAD + ∠AOB)
⇒ ∠ADB = 360° - (90° + 90° + 128°)
⇒ ∠ADB = 360° - 308°
⇒ ⇒ ∠ADB = 52°
∴ The required measure of ∠ADB is 52°.
The radius of the circle 2x2 + 2y2 + 8x + 8y + 4 = 0 is
Answer (Detailed Solution Below)
Circles Question 7 Detailed Solution
Download Solution PDFConcept:
The general second degree equation of a circle in x and y is given by: x2 + y2 + 2gx + 2fy + c = 0 with centre (-g, -f) and radius
Calculation:
Given: 2x2 + 2y2 + 8x + 8y + 4 = 0
⇒ 2 × (x2 + y2 + 4x + 4y + 2) = 0
⇒ x2 + y2 + 4x + 4y + 2 = 0 are equation of circle with centres C and radius r
By comparing the equation of the circle with the equation x 2 + y2 + 2gx + 2fy + c = 0 we get
g = 2, f = 2 and c = 2
As we know, radius =
r = √6 unit
In a circle with centre O, a 6 cm long chord is at a distance 4 cm from the centre. Find the length of the diameter.
Answer (Detailed Solution Below)
Circles Question 8 Detailed Solution
Download Solution PDFConcept:
Any line from the center that bisects a chord is perpendicular to the chord.
Pythagoras theorem:
h2 = b2 + p2
Calculation:
r2 = 42 + 32
⇒ r = 5
∴ The diameter of the circle = 2r = 2 × 5
⇒ 10 cm
Shortcut TrickBy triplets:
3, 4 and 5
The length of the diameter = 2× 5 = 10
The circle
Answer (Detailed Solution Below)
Circles Question 9 Detailed Solution
Download Solution PDFConcept:
On X- axis y-intercept will be zero, similarly, on Y-axis x-intercept will be zero.
Calculation:
On Y-axis x-intercept = 0
So,
y = 3 and 4
Points (0, 3) and (0, 4)
Now length =
Hence, option (4) is correct.
AB is the diameter of a circle with centre O. C and D are two points on the circle on either side of AB, such that ∠CAB = 52° and ∠ABD = 47°. What is the difference (in degrees) between the measures of ∠CAD and ∠CBD?
Answer (Detailed Solution Below)
Circles Question 10 Detailed Solution
Download Solution PDFGiven:
∠CAB = 52° and ∠ABD = 47°
Concept used:
The angle subtended by the diameter at any point on the circumference is equal to 90°
Calculation:
According to the concept,
∠ACB = ∠ADB = 90°
So, ∠CBA = 180° - (90° + 52°)
⇒ 180° - 142°
⇒ 38°
Similarly,
∠BAD = 180° - (90° + 47°)
⇒ 180° - 137°
⇒ 43°
So, ∠CAD = 52° + 43°
⇒ 95°
∠CBD = 38° + 47°
⇒ 85°
Now,
∠CAD - ∠CBD = 95° - 85°
⇒ 10°
∴ The difference (in degrees) between the measures of ∠CAD and ∠CBD is 10.
The sum of the lengths of the radius and the diameter of a circle is 84 cm. What is the difference between the lengths of the circumference and the radius of this circle? [Use
Answer (Detailed Solution Below)
Circles Question 11 Detailed Solution
Download Solution PDFGiven:
Radius + Diameter = 84 cm
Formula used:
Diameter = 2 x Radius
Circumference = 2πr
Calculations:
Let the radius be R then, Diameter be 2R.
Now, according to question,
R + 2R = 84
⇒ 3R = 84
⇒ R = 28
So, The circumference = 2πr = 2 ×
Now, difference between Circumference and radius is
= 176 - 28
= 148 cm
Hence, the required difference is 148 cm.
What is the length of the tangent to the circle x2 + y2 = 9 from the point (4, 0).
Answer (Detailed Solution Below)
Circles Question 12 Detailed Solution
Download Solution PDFConcept:
The length of the tangent from an external point (x1, y1) to the circle x2 + y2 = a2 is
Calculation:
Given: Equation of circle x2 + y2 = 9 and the point (4, 0).
As we know that, the length of the tangent from an external point (x1, y1) to the circle x2 + y2 = a2 is
Here, x1 = 4 , y1 = 0 and a2 = 9.
So, the length of the tangent is √7 units.
Find the equation of the circle whose centre is at (2, - 3) and which passes through the intersection of the lines 3x + 2y = 11 and 2x + 3y = 4 ?
Answer (Detailed Solution Below)
Circles Question 13 Detailed Solution
Download Solution PDFCONCEPT:
Equation of circle with centre at (h, k) and radius r units is given by: (x - h)2 + (y - k)2 = r2
CALCULATION:
Here, we have to find the equation of the circle whose centre is at (2, - 3) and which passes through the intersection of the lines 3x + 2y = 11 and 2x + 3y = 4
First let's find the point of intersection of the lines 3x + 2y = 11 and 2x + 3y = 4
So, by solving the equations 3x + 2y = 11 and 2x + 3y = 4, we get x = 5 and y = - 2
So, the required circle passes through the point (5, - 2)
Let the radius of the required circle be r
As we know that, the equation of circle with centre at (h, k) and radius r units is given by: (x - h)2 + (y - k)2 = r2
Here, we have h = 2 and k = - 3
⇒ (x - 2)2 + (y + 3)2 = r2 ------------(1)
∵ The circle passes through the point (5, - 2)
So, x = 5 and y = - 2 will satisfy the equation (1)
⇒ (5 - 2)2 + (- 2 + 3)2 = r2
⇒ r2 = 10
So, the equation of the required circle is (x - 2)2 + (y + 3)2 = 10
⇒ x2 + y2 - 4x + 6y + 3 = 0
So, the equation of the required circle is x2 + y2 - 4x + 6y + 3 = 0
Hence, option A is the correct answer.
The equation of circle with centre (1, -2) and radius 4 cm is:
Answer (Detailed Solution Below)
Circles Question 14 Detailed Solution
Download Solution PDFGiven
Centre point are (1, -2)
Radius = 4cm
Formula used
(x -a)2 + (y - b)2 = r2
where, a and b are point on centre
r = radius
x and y be any point on circle
Calculation
Put the value of a, b and r in the formula
(x-1)2 + (y + 2)2 = 16
⇒ x2 + 1 - 2x + y2 + 4 + 4y = 16
⇒ x2 + y2 - 2x + 4y = 11
The radius of the circle x2 + y2 + x + c = 0 passing through the origin is
Answer (Detailed Solution Below)
Circles Question 15 Detailed Solution
Download Solution PDFConcept:
Let x2 + y2 = r2 is the equation of circle. then (0, 0) is the origin and r is the radius of the circle.
Calculations:
We know that, x2 + y2 = r2 is the equation of circle. then (0, 0) is the origin and r is the radius of the circle.
Given equation of circle is x2 + y2 + x + c = 0, which is passing through the origin.
i.e. c = 0.
⇒x2 + y2 + x = 0.
⇒ x2 + x +
⇒x2 + x +
⇒
which is equation of circle with radius is
Hence, the radius of the circle x2 + y2 + x + c = 0 passing through the origin is