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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.
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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 x^{2} + y^{2} + 4x – 6y = 5 is:

Solution:

x^{2}+y^{2}+4x-6y=5

Circle Equation

(x-a)^{2}+(y-b)^{2}=r^{2 }is the circle equation with a radius r, centered at (a,b)

Rewrite^{ }x^{2}+y^{2}4x-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 x^{2} + y^{2} - 4x - 6y - 3 = 0 and which touches the X-axis is:

Solution:

**x² + y² -4x -6y -3 = 0**

** (x² -4x +4 -4) + (y² -6y +9-9) = 3**

**(x²-4x +4) +(y²-6y+9) = 3+4+9**

**(x-2) ² +(y-3) ²=16**

**This is circle with center(2,3) and radius 4**

**Since required circle is concentric to this circle**

**So its center will be (2,3)**

**Let circle be**

**(x-2) ²+(y-3) ²=r²…………(1)**

**Now we need radius of this circle.**

**As we are also given that circle touches y axis**

**So y axis is tangent to the circle.**

**Point of contact will be (0,3)**

**Let’s substitute (0,3) in circle (1)**

**(0-2) ² + (3-3) ²=r²**

**So radius is 2 as r²=4**

**So eqn of the circle becomes**

**(x-2) ²+(y-3) ²=4**

**x² +y² -4x-6x+4+9=4**

**x² +y² -4x-6x+9=0**

QUESTION: 10

Point (-2, – 5) lies on the circle x^{2} + y^{2} = 25.

Solution:

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