Test: Circles -1


10 Questions MCQ Test Mathematics (Maths) Class 11 | Test: Circles -1


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QUESTION: 1

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

Solution:

QUESTION: 2

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

Solution:


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:

QUESTION: 6

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

Solution:
QUESTION: 7

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

Solution:

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:
QUESTION: 10

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

Solution:

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