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# Test: Circles- 1

## 10 Questions MCQ Test Mathematics For JEE | Test: Circles- 1

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This mock test of Test: Circles- 1 for JEE helps you for every JEE entrance exam. This contains 10 Multiple Choice Questions for JEE Test: Circles- 1 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Circles- 1 quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Test: Circles- 1 exercise for a better result in the exam. You can find other Test: Circles- 1 extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

### The equation of parabola whose focus is (– 3, 0) and directrix x + 5 = 0 is:

Solution:

Given, focus : (-3,0)
directrix : x + 5 = 0
Let (x ,y) is the point on the parabola .
∴ distance of point from focus = distance of point from directrix
⇒ √{(x + 3)² + y²} = |x + 5|/√(1² + 0²)
⇒ √{(x + 3)² + y² } = |x + 5|
squaring both sides,
(x + 3)² + y² = (x + 5)²
⇒y² = (x + 5)² - (x + 3)²
⇒y² = (x + 5 - x - 3)(x + 5 + x + 3)
⇒y² = 2(2x + 8) = 4(x + 4)
Hence, equation of parabola is y² = 4(x + 4)

QUESTION: 2

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

Solution:

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

QUESTION: 3

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

Solution:

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

QUESTION: 4

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

Solution:

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

QUESTION: 5

The equation of a circle with centre as the origin is

Solution:

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

QUESTION: 6

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

Solution:

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

QUESTION: 7

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

Solution:

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

QUESTION: 8

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

Solution:

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 QUESTION: 9

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

Solution:

x2 + y2 - 4x - 6y + 4 = 0

QUESTION: 10

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

Solution:

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