Vidaara.orgClass 11 · Mathematics
CodeVID-M11-10-CIR-01
Circle — Assignment
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Show all working for Sections B, C and D. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
The circle with centre origin and radius $3$ is:
- A.$x^2+y^2=3$
- B.$x^2+y^2=9$
- C.$x^2+y^2=6$
- D.$x+y=3$
2.
The centre of $(x-1)^2+(y-2)^2=9$ is:
- A.$(1,2)$
- B.$(-1,-2)$
- C.$(2,1)$
- D.$(1,-2)$
3.
The radius of $x^2+y^2=49$ is:
- A.$49$
- B.$7$
- C.$\sqrt7$
- D.$14$
4.
The centre of $x^2+y^2+2gx+2fy+c=0$ is:
- A.$(g,f)$
- B.$(-g,-f)$
- C.$(c,c)$
- D.$(0,0)$
5.
The radius of $x^2+y^2-4x=0$ is:
- A.$1$
- B.$2$
- C.$4$
- D.$\sqrt2$
Section B — Short Answer (2 marks)
4 × 2 = 8 marks
6.
Write the equation of the circle with centre origin and radius $5$.
7.
Find the centre and radius of $(x-3)^2+(y+1)^2=16$.
8.
Find the equation of the circle with centre $(1,2)$ and radius $3$.
9.
Find the radius of $x^2+y^2-6x-8y=0$.
Section C — Short Answer (3 marks)
4 × 3 = 12 marks
10.
Find the centre and radius of $x^2+y^2-4x-6y-12=0$.
11.
Find the equation of the circle with centre $(2,-3)$ passing through the origin.
12.
Find the equation of the circle with the segment joining $(1,2)$ and $(3,4)$ as diameter.
13.
Does the point $(3,4)$ lie on $x^2+y^2=25$?
Section D — Long Answer (5 marks)
2 × 5 = 10 marks
14.
Find the equation of the circle passing through $(0,0),(4,0)$ and $(0,6)$.
15.
Find the centre and radius of $2x^2+2y^2-8x+12y-1=0$.
Answer Key
Section A — Multiple Choice Questions
- (B) $x^2+y^2=9$
- (A) $(1,2)$
- (B) $7$
- (B) $(-g,-f)$
- (B) $2$
Section B — Short Answer (2 marks)
- $x^2+y^2=25$.
- Centre $(3,-1)$, radius $4$.
- $(x-1)^2+(y-2)^2=9$.
- $5$ (centre $(3,4)$).
Section C — Short Answer (3 marks)
- Centre $(2,3)$, radius $5$.
- $(x-2)^2+(y+3)^2=13$.
- $(x-2)^2+(y-3)^2=2$.
- Yes, it lies on the circle.
Section D — Long Answer (5 marks)
- $x^2+y^2-4x-6y=0$.
- Centre $(2,-3)$, radius $\dfrac{3\sqrt6}{2}$.
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