Test: Circle- 1 - Commerce MCQ

# Test: Circle- 1 - Commerce MCQ

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## 10 Questions MCQ Test Mathematics (Maths) Class 11 - Test: Circle- 1

Test: Circle- 1 for Commerce 2024 is part of Mathematics (Maths) Class 11 preparation. The Test: Circle- 1 questions and answers have been prepared according to the Commerce exam syllabus.The Test: Circle- 1 MCQs are made for Commerce 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Circle- 1 below.
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Test: Circle- 1 - Question 1

### The equation of the circle passing through (0, 0) and making intercepts 2 and 4 on the coordinate axes is:

Detailed Solution for Test: Circle- 1 - Question 1

The circle intercept the co-ordinate axes at a and b. it means x - intercept at ( a, 0) and y-intercept at (0, b) .
Now, we observed that circle passes through points (0, 0) , (a, 0) and (0, b) .
we also know, General equation of circle is
x² + y² + 2gx + 2fy + C = 0
when point (0,0)
(0)² + (0)² + 2g(0) + 2f(0) + C = 0
0 + 0 + 0 + 0 + C = 0
C = 0 -------(1)
when point (a,0)
(a)² + (0)² + 2g(a) + 2f(0) + C = 0
a² + 2ag + C = 0
from equation (1)
a² + 2ag = 0
a(a + 2g) = 0
g = -a/2
when point ( 0, b)
(0)² + (b)² + 2g(0) + 2f(b) + C = 0
b² + 2fb + C = 0
f = -b/2
Now, equation of circle is
x² + y² + 2x(-a/2) + 2y(-b/2) + 0 = 0 { after putting values of g, f and C }
x² + y² - ax - by = 0
As we know that, a=2, b=4
x^2 + y^2 - 2x - 4y = 0

Test: Circle- 1 - Question 2

### The equation of a circle with centre as the origin is

Detailed Solution for Test: Circle- 1 - Question 2

The equation of a circle with center (h,k) and radius r is given by (x−h)2+(y−k)2=r2
For a circle centered at the origin, this becomes the more familiar equation x2+y2=r2

Test: Circle- 1 - Question 3

### A circle is the set of …… in a plane that are equidistant from a fixed point in the plane.

Detailed Solution for Test: Circle- 1 - Question 3

A circle is the set of points a fixed distance from a center point,

Test: Circle- 1 - Question 4

The centre and radius of the circle x2 + y2 + 4x – 6y = 5 is:

Detailed Solution for Test: Circle- 1 - Question 4

x2+y2+4x-6y=5
Circle Equation
(x-a)2+(y-b)2=r2 is the circle equation with a radius r, centered at (a,b)
Rewrite x2+y24x-6y=5 in the form of circle standard circle equation
(x-(-2))2+(y-3)2=(3√2)2
Therefore the circle properties are:

(a,b) = (-2,3), r = 3√2

Test: Circle- 1 - Question 5

The equation of circle whose centre is (2, 1) and which passes through the point (3, – 5) is:

Detailed Solution for Test: Circle- 1 - Question 5

Radius of circle is given by -
r = √[(h-x1)² + (k-y1)²]
r = √[(2-3)² + (1+5)²]
r = √(-1² + 6²)
r = √(1 + 36)
r = √37
if centre (2,-]1) and radius=√26 are given,
(x-h)2+(y-k)2=r2
equation is (x-2)2 + (y-1)2 = (√37)2
x2 + 4 - 4x + y2 + 1 - 2y = 37
x2 + y2 - 4x - 2y - 32 = 0

Test: Circle- 1 - Question 6

The equation of circle of radius 5 units touches the coordinates axes in the second quadrant is:

Detailed Solution for Test: Circle- 1 - Question 6

If the circle lies in second quadrant
The equation of a circle touches both the coordinate axes and has radius a is
x2 + y2 + 2ax - 2ay + a2 = 0
Radius of circle, a = 5
x2 + y2 + 10x - 10y + 25 = 0

Test: Circle- 1 - Question 7

The equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes is:

Detailed Solution for Test: Circle- 1 - Question 7

Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.
Since the centre of the circle passes through (0, 0),
(0 – h)2 + (0 – k)2 = r2
⇒ h2 + k2 = r2
The equation of the circle now becomes (x – h)2 + (y – k)2 = h2 + k2.
It is given that the circle makes intercepts a and b on the coordinate axes. This means that the circle passes through points (a, 0) and (0, b). Therefore,
(a – h)2 + (0 – k)2 = h2 + k2 …………… (1)
(0 – h)2 + (b – k)2 = h2 + k2 ………… (2)
From equation (1), we obtain a2 – 2ah + h2 + k2 = h2 + k2
⇒ a2 – 2ah = 0
⇒ a(a – 2h) = 0
⇒ a = 0 or (a – 2h) = 0
However, a ≠ 0; hence, (a – 2h) = 0 ⇒ h =a/2.
From equation (2), we obtain h2 + b2 – 2bk + k2 = h2 + k2
⇒ b2 – 2bk = 0
⇒ b(b – 2k) = 0
⇒ b = 0 or(b – 2k) = 0
However, b ≠ 0; hence, (b – 2k) = 0 ⇒ k =b/2. Thus, the equation of the required circle is

Test: Circle- 1 - Question 8

The equation of a circle which is concentric to the given circle x2 + y2 - 4x - 6y - 3 = 0 and which touches the Y-axis is:

Detailed Solution for Test: Circle- 1 - Question 8

Test: Circle- 1 - Question 9

Point (-2, – 5) lies on the circle x2 + y2 = 25.

Detailed Solution for Test: Circle- 1 - Question 9

X2 +Y2 =25    Points =(-2,-5)
Put the points in the variables
(-2)2 + (-5)2 = 29
As 29 > 25  (lies outside).

Test: Circle- 1 - Question 10

The locus of the centre of a circle which passes through the point (a, 0) and touches the line x+1=0, is

Detailed Solution for Test: Circle- 1 - Question 10
The locus is a parabola as distance of centre from straight line = its distance from point (a, 0).

## Mathematics (Maths) Class 11

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## Mathematics (Maths) Class 11

75 videos|238 docs|91 tests