Circles Test 10

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Circles Test 10

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Weekly Quiz Competition

Question 1
1 / -0

If $$(a,b)$$ lies on circle with centre as origin, then its radius will be

A
$$\displaystyle a-b$$

B
$$\displaystyle a+b$$

C
$$\displaystyle \sqrt { { a }^{ 2 }+{ b }^{ 2 } } $$

D
$$\displaystyle { a }^{ 2 }+{ b }^{ 2 }$$

Solution

We know the formula,

The equation of a circle of radius r and centre the origin is

$$x^2+y^2=r^2$$

Here the center is (a, b)

so Radius , $$r = \sqrt{a^2+b^2}$$
Question 2
1 / -0

Which of the following is an equation of the circle with its center at $$(0,0)$$ that passes through $$(3,4)$$ in the standard $$(x,y)$$ coordinate plane?

A
$$x+y=1$$

B
$$x-y=25$$

C
$${x}^{2}+y=25$$

D
$${x}^{2}+{y}^{2}=5$$

E
$${x}^{2}+{y}^{2}=25$$

Solution

Let the radius of the circle be '$$r$$'
Then the equation of the circle is
$$(x-0)^{2}+(y-0)^{2}=r^{2}$$
Hence $$x^{2}+y^{2}=r^{2}$$
Now, it passes through $$(3,4)$$.
Hence $$3^{2}+4^{2}=r^{2}$$ $$\Rightarrow r^{2}=25$$.
Therefore, the equation of the circle is $$x^{2}+y^{2}=25$$.
Question 3
1 / -0

A circle with center $$(3, 8)$$ contains the point $$(2, -1)$$. Another point on the circle is:

A
$$(1, -10)$$

B
$$(4, 17)$$

C
$$(5, -9)$$

D
$$(7, 15)$$

E
$$(9, 6)$$

Solution

$$(x-3)^2+(y-8)^2=r^2$$
Given $$(2,-1)$$ lies on the circle
$$(2-3)^2+(-1-8)^2=r^2$$
$$1+81=r^2$$
$$r^2=82$$
Circle equation is :$$(x-3)^2+(y-8)^2=82$$
By trial and error, substitute the point
in the above equation
$$(4-3)^2+(17-8)^2=82$$
hence , $$(4,17)$$ satisfy the circle euation.
Question 4
1 / -0

If the centre $$O$$ of circle is the intersection of $$x-$$axis and line $$y=\dfrac { 4 }{ 3 } x+4$$, and the point $$(3,8)$$ lies on circle, then the equation of circle will be

A
$$\displaystyle { x }^{ 2 }+{ y }^{ 2 }=25$$

B
$$\displaystyle { \left( x+3 \right) }^{ 2 }+{ y }^{ 2 }=25$$

C
$$\displaystyle { \left( x+3 \right) }^{ 2 }+{ y }^{ 2 }=100$$

D
$$\displaystyle { \left( x+3 \right) }^{ 2 }+{ \left( y-8 \right) }^{ 2 }=100$$

Solution

The intersection of $$x$$-axis and the line $$y = \cfrac{4x}{3} + 4$$ is $$(-3,0)$$, which is the centre of the circle.
Since the point $$(3,8)$$ lies on the circle, its radius becomes the distance between these points, which is $$\sqrt{(3 + 3)^2 + (8 - 0)^2} = 10$$
The equation thus becomes $$(x + 3)^2 + (y - 0)^2 = (10)^2$$
i.e. $$(x + 3)^2 + y^2 = 100$$
Question 5
1 / -0

The area of the circle represented by the equation $${(x+3)}^{2}+{(y+1)}^{2}=25$$ is

A
$$4\pi$$

B
$$5\pi$$

C
$$16\pi$$

D
$$25\pi$$

Solution

The general equation of circle with centre $$(h,k)$$ and radius $$r$$ is given by $${ \left( x-h \right) }^{ 2 }+{ \left( y-k \right) }^{ 2 }={ \left( r \right) }^{ 2 }$$.
It is given that the equation of the circle is $$(x+3)^2+(y+1)^2=25$$
Therefore, $$h=-3$$, $$k=-1$$ and $$r=5$$.
The area of the circle with radius $$r$$ is $$\pi { r }^{ 2 }$$, thus the area of the circle with radius $$r=5$$ is:
$$\pi { r }^{ 2 }=\pi { (5) }^{ 2 }=25\pi$$
Hence, the area of the circle is $$25\pi$$.
Question 6
1 / -0

Which of the following is an equation of a circle in the $$xy$$-plane with center $$\left(0, 4\right)$$ and a radius with endpoint $$\left(\dfrac{4}{3}, 5\right)$$?

A
$${ x }^{ 2 }+{ \left( y-4 \right) }^{ 2 }=\dfrac { 25 }{ 9 } $$

B
$${ x }^{ 2 }+{ \left( y+4 \right) }^{ 2 }=\dfrac { 25 }{ 9 } $$

C
$${ x }^{ 2 }+{ \left( y-4 \right) }^{ 2 }=\dfrac { 5 }{ 3 } $$

D
$${ x }^{ 2 }+{ \left( y+4 \right) }^{ 2 }=\dfrac { 3 }{ 5 } $$

Solution

Given, center of circle is $$(0,4)$$ and point on circle $$\left (\dfrac 43,5\right)$$
$$\therefore$$ radius of circle $$=\sqrt {\left (\dfrac 43-0\right)^2+(5-4)^2}$$
$$= \sqrt {\dfrac {16}{9}+1}=\sqrt {\dfrac {25}9}$$
Since, Equation of the circle having centre $$(x_1,y_1)$$ and radius '$$r$$' is
$$(x-x_1)^2+(y-y_1)^2=r^2$$
$$\therefore$$ Equation of given circle $$= (x-0)^2+(y-4)^2=\left (\sqrt {\dfrac {25}9}\right)^2$$
$$= x^2+(y-4)^2=\dfrac {25}9$$
Hence, option A is correct.
Question 7
1 / -0

The lines $$2x - 3y - 5 = 0$$ and $$3x -4y = 7$$ are diameters of a circle of area 154 sq units, then the equation of the circle is.( Use $$\pi = \dfrac{22}{7}$$)

A
$$x^2+y^2+2x-2y-62=0$$

B
$$x^2+y^2+2x-2y-47=0$$

C
$$x^2+y^2-2x+2y-47=0$$

D
$$x^2+y^2-2x-2y-62=0$$

Solution

The centre of the required circle lies at the intersection of $$2x - 3y- 5 = 0$$ and $$3x - 4y - 7 = 0$$.

Thus, the coordinates of the centre are $$(1, -1)$$.

Let $$r$$ be the radius of the circle.

$$\pi r^2 =154 $$

$$\Rightarrow \dfrac{22}{7}r^2=154 \Rightarrow r=7$$

Hence, the equation of required circle is $$(x -1)^2 +(y+ 1)^2= 7^2$$

$$\Rightarrow x^2+y^2-2x+2y-47=0$$
Question 8
1 / -0

The least value of $$2x^{2} + y^{2} + 2xy + 2x - 3y + 8$$ for real numbers $$x$$ and $$y$$ is

A
$$2$$

B
$$8$$

C
$$3$$

D
$$1$$

E
$$-\dfrac12$$

Solution

$$E = 2x^{2} + y^{2} + 2xy + 2x - 3y + 8$$

$$\Rightarrow E = \dfrac {1}{2} (4x^{2} + 2y^{2} + 4xy + 4x - 6y + 16)$$

$$\Rightarrow E= \dfrac {1}{2}\left \{(y - 4)^{2} + (2x + y + 1)^{2} - 1\right \} \geq - \dfrac {1}{2}$$

This happen when the $$\left \{(y - 4)^{2} + (2x + y + 1)^{2} - 1\right \} = -1$$
Which i possible when $$(x,y) \equiv \left(-\dfrac52,4\right)$$

So, least value is $$-\dfrac {1}{2}$$
Question 9
1 / -0

Let $$ABCD$$ be a square of side length $$1$$. and $$\Gamma $$ a circle passing through $$B$$ and $$C$$, and touching $$AD$$. The radius of $$\Gamma $$ is

A
$$\cfrac { 3 }{ 8 } $$

B
$$\cfrac { 1 }{ 2 } $$

C
$$\cfrac { 1 }{ \sqrt { 2 } } $$

D
$$\cfrac { 5 }{ 8 } $$

Solution

$$PC=r$$
$${ PC }^{ 2 }={ r }^{ 2 }$$
$${ \left( r-1 \right) }^{ 2 }+{ \left( \cfrac { 1 }{ 2 } -1 \right) }^{ 2 }={ r }^{ 2 }$$
$$1-2r+\cfrac { 1 }{ 4 } =0$$
$$r=\cfrac { 5 }{ 8 } $$
Question 10
1 / -0

Consider the parametric equation
$$x = \dfrac {a(1 - t^{2})}{1 + t^{2}}, y = \dfrac {2at}{1 + t^{2}}$$.
What does the equation represent?

A
It represents a circle of diameter $$a$$

B
It represents a circle of radius $$a$$

C
It represents a parabola

D
None of the above

Solution

$$x=\dfrac { a(1-{ t }^{ 2 }) }{ 1+{ t }^{ 2 } }, y=\dfrac { 2at }{ 1+{ t }^{ 2 } }$$
Let $$t=\tan\theta$$
$$\Rightarrow x=a\sin 2\theta , y=a\cos 2\theta \\ \Rightarrow { x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }$$
It represents a circle of radius $$a$$.

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Re-Attempt Weekly Quiz Competition

Circles Test 10

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Circles Test 10

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SHARING IS CARING

Question 1

$$\displaystyle a-b$$

$$\displaystyle a+b$$

$$\displaystyle \sqrt { { a }^{ 2 }+{ b }^{ 2 } } $$

$$\displaystyle { a }^{ 2 }+{ b }^{ 2 }$$

Solution

Question 2

$$x+y=1$$

$$x-y=25$$

$${x}^{2}+y=25$$

$${x}^{2}+{y}^{2}=5$$

$${x}^{2}+{y}^{2}=25$$

Solution

Question 3

$$(1, -10)$$

$$(4, 17)$$

$$(5, -9)$$

$$(7, 15)$$

$$(9, 6)$$

Solution

Question 4

$$\displaystyle { x }^{ 2 }+{ y }^{ 2 }=25$$

$$\displaystyle { \left( x+3 \right) }^{ 2 }+{ y }^{ 2 }=25$$

$$\displaystyle { \left( x+3 \right) }^{ 2 }+{ y }^{ 2 }=100$$

$$\displaystyle { \left( x+3 \right) }^{ 2 }+{ \left( y-8 \right) }^{ 2 }=100$$

Solution

Question 5

$$4\pi$$

$$5\pi$$

$$16\pi$$

$$25\pi$$

Solution

Question 6

$${ x }^{ 2 }+{ \left( y-4 \right) }^{ 2 }=\dfrac { 25 }{ 9 } $$

$${ x }^{ 2 }+{ \left( y+4 \right) }^{ 2 }=\dfrac { 25 }{ 9 } $$

$${ x }^{ 2 }+{ \left( y-4 \right) }^{ 2 }=\dfrac { 5 }{ 3 } $$

$${ x }^{ 2 }+{ \left( y+4 \right) }^{ 2 }=\dfrac { 3 }{ 5 } $$

Solution

Question 7

$$x^2+y^2+2x-2y-62=0$$

$$x^2+y^2+2x-2y-47=0$$

$$x^2+y^2-2x+2y-47=0$$

$$x^2+y^2-2x-2y-62=0$$

Solution

Question 8

$$2$$

$$8$$

$$3$$

$$1$$

$$-\dfrac12$$

Solution

Question 9

$$\cfrac { 3 }{ 8 } $$

$$\cfrac { 1 }{ 2 } $$

$$\cfrac { 1 }{ \sqrt { 2 } } $$

$$\cfrac { 5 }{ 8 } $$

Solution

Question 10

It represents a circle of diameter $$a$$

It represents a circle of radius $$a$$

It represents a parabola

None of the above

Solution

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