Circles Test 1

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Circles Test 1

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Weekly Quiz Competition

Question 1
1 / -0

Let $$C$$ be the circle with centre at $$(1,\ 1)$$ and radius $$=1$$. If $$T$$ is the circle centered at $$ (0,\ y)$$, passing through origin and touching the circle $$C$$ externally, then the radius of $$T$$ is equal to :

A
$$ \displaystyle \frac{\sqrt{3}}{\sqrt{2}}$$

B
$$ \displaystyle \frac{\sqrt{3}}{2}$$

C
$$ \displaystyle \frac{1}{2}$$

D
$$ \displaystyle \frac{1}{4}$$

Solution

Distance between there centers,
$$\Rightarrow$$ $$k+1=\sqrt{1+(k-1)^2}$$
$$\Rightarrow$$ $$k+1=\sqrt{1+k^2+1-2k}$$
$$\Rightarrow$$ $$k+1=\sqrt{k^2+2-2k}$$
$$\Rightarrow$$ $$(k+1)^2=k^2+2-2k$$
$$\Rightarrow$$ $$k^2+2k+1=k^2+2-2k$$
$$\Rightarrow$$ $$4k=1$$
$$\therefore$$ $$k=\dfrac{1}{4}$$
Question 2
1 / -0

If the lines $$2\mathrm{x}+3\mathrm{y}+1=0$$ and $$\mathrm{3x}- \mathrm{y}-4=0$$ lie along diameters of a circle of circumference $$ 10\pi$$, then the equation of the circle is:

A
$$\mathrm{x}^{2}+\mathrm{y}^{2}-2\mathrm{x}+2\mathrm{y}-23=0$$

B
$$\mathrm{x}^{2}+\mathrm{y}^{2}-2\mathrm{x}-2\mathrm{y}-23=0$$

C
$$\mathrm{x}^{2}+\mathrm{y}^{2}+2\mathrm{x}+2\mathrm{y}-23=0$$

D
$$\mathrm{x}^{2}+\mathrm{y}^{2}+2\mathrm{x}-2\mathrm{y}-23=0$$

Solution

Two diameters are along

$$2x+3y+1=0$$

$$3x−y−4=0$$

$$2x+3y+1=0$$ ..............$$(1)$$

$$3x−y−4=0$$ ...............$$(2) \times 3$$

Subtract $$(2)-(1)$$

$$\Rightarrow x =1$$ and $$y=-1$$

on solving we get center as $$(1,−1)$$

We know that circumference$$ =2\pi r$$

$$2\pi r=10\pi $$

$$2r=10$$

$$r=5$$

Required circle is $$(x−1)^2+(y+1)^2=52$$

$$x^2+y^2−2x+2y−23=0$$
Question 3
1 / -0

The circle $$x^{2}+y^{2}-8x=0$$ and hyperbola $$\dfrac{x^{2}}{9}-\dfrac{y^{2}}{4}=1$$ intersect at the points $$A$$ and $$B$$.
then the equation of the circle with $$AB$$ as its diameter is

A
$$x^{2}+y^{2}-12x+24=0$$

B
$$x^{2}+y^{2}+12x+24=0$$

C
$$x^{2}+y^{2}+24x-12=0$$

D
$$x^{2}+y^{2}-24x-12=0$$

Solution

Given: $$x^2+y^2-8x=0 .....(1)$$ $$\implies y^2=8x-x^2$$

$$\dfrac {x^2}{9}-\dfrac {y^2}{4}\Rightarrow 1$$

$$\Rightarrow\dfrac {x^2}{9}-\dfrac {(8x-x^2)}{4}=1$$

$$\Rightarrow4x^2-9(8x-x^2)=36$$

$$\Rightarrow4x^2-72x+9x^2-36=0$$

$$\Rightarrow13x^2-72x-36=0$$

$$\Rightarrow x=\dfrac {72\pm \sqrt {(72)^2-4\times 13\times 36}}{26}$$

$$\Rightarrow x=6$$

Then, putting $$x=6$$ in equation (1), we get

$$\Rightarrow y=\pm \sqrt{12}=\pm \sqrt{4\times3}=\pm 2\sqrt 3$$

So, radius $$=2\sqrt 3$$

As $$AB$$ is a diameter center will be the midpoint of point $$A=(6,2\sqrt 3)$$ and $$B=(6,-2\sqrt 3)$$

By mid point formula,
Center $$=\left(\dfrac{6+6}{2},\dfrac{2\sqrt 3-2\sqrt 3}{2}\right)$$

Centre $$=(6, 0)$$

$$\Rightarrow(x-6)^2+y^2=12$$

$$\Rightarrow x^2+36-12x+y^2=12$$

$$\Rightarrow x^2-12x+y^2+24=0$$
Question 4
1 / -0

The equation of the circle touching $$x = 0, y = 0$$ and $$x = 4$$ is

A
$$x^2 + y^2 - 4x - 4y + 16 = 0$$

B
$$x^2 + y^2 - 8x - 8y + 16 = 0$$

C
$$x^2 + y^2 + 4x + 4y + 4 = 0$$

D
$$x^2 + y^2 - 4x - 4y + 4 = 0$$

Solution

The given circle is as shown in the following figure
It is clear that the coordinates of the centre of the circle are (2, 2) and radius = 2
Hence, the required equation of the circle is $$(x - 2)^2 + (y - 2)^2 = (2)^2$$
$$\Rightarrow x^2 + y^2 - 4x - 4y + 4 = 0$$
Question 5
1 / -0

The radius of the circle with center (0,0) and which passes through (-6,8) is

A
5

B
10

C
6

D
8

Solution

$$\displaystyle r=\sqrt{\left ( 6 \right )^{2}+\left ( -8 \right )}=10$$
Question 6
1 / -0

The equation of circle with its centre at the origin is $$x^2+y^2=r^2$$

A
True

B
False

C
Neither

D
Either

Solution

The center is a fixed point in the middle of the circle; usually given the general coordinates (h, k).

The fixed distance from the center to any point on the circle is called the radius.

A circle can be represented by two different forms of equations, the general form and the center-radius form. This discussion will focus on the center-radius form.
centre radius form
let (h,k) be the centre of a cicle and r be the radius
$${(x-h)}^{2}+{(y-k)}^{2}={r}^{2}$$
and we talk about centre at origin
than $$(h,k)$$ will become $$(0,0)$$
so, equation will become $${x}^{2}+{y}^{2}={r}^{2}$$
therefore option A will be answer.
Question 7
1 / -0

Which of the following equations of a circle has center at (1, -3) and radius of 5?

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

B
$$\displaystyle \left ( x-1 \right )^{2}+\left ( y+3 \right )^{2}=25$$

C
$$\displaystyle \left ( x-1 \right )^{2}+\left ( y-3 \right )^{2}=25$$

D
$$\displaystyle \left ( x+1 \right )^{2}+\left ( y-3 \right )^{2}=25$$

Solution

the general equation of a circle with center at (a,b) and radius r is
$$(x-a)^2+(y-b)^2=r^2$$
so substituting the values we get the circle equation as
$$(x-1)^2+(y+3)^2=25$$ option B
Question 8
1 / -0

Determine the area enclosed by the curve $$\displaystyle x^{2}-10x+4y+y^{2}=196$$

A
$$15\pi$$

B
$$225\pi$$

C
$$20\pi$$

D
$$17\pi$$

Solution

Equation of the circle is
$$x^2-10x+4y+y^2=196$$
$$=>x^2-2\times x \times 5 +5^2+y^2+2\times y\times 2+2^2=196+2^2+5^2$$
$$=>(x-5)^2+(y+2)^2=196+4+25$$
$$=>(x-5)^2+(y+2)^2=225$$
$$=>(x-5)^2+(y+2)^2=15^2$$
Thus radius of the circle is 15
Thus, area of the circle $$=\pi \times 15^2$$
$$=225\pi$$
Question 9
1 / -0

The standard equation of circle at origin is

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

B
$$x^2-y^2=r$$

C
$$x^2+y^2=1$$

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

Solution

The standard equation of circle at origin is $$x^2+y^2=r^2$$
Example: A circle with its center at the origin $$(0, 0)$$, and radius $$r$$ has the equation $$x^2 + y^2 = r^2$$. We'll know if a given point is on the circle if the coordinates of that point satisfy the equation.
Question 10
1 / -0

The diameter of a circle described by $$\displaystyle 9x^{2}+9y^{2}=16$$ is

A
$$\dfrac {16}9$$

B
$$\dfrac 43$$

C
$$4$$

D
$$\dfrac 83$$

Solution

Equation of circle is
$$9x^2+9y^2=16$$
$$=>x^2+y^2=\dfrac{16}{9}$$
$$=>x^2+y^2=(\dfrac{4}{3})^2$$
$$\therefore$$ Radius of the circle =$$\dfrac{4}{3}$$
$$\therefore$$ Diameter of the circle=$$\dfrac{4\times 2}{3}$$
$$=\dfrac{8}{3}$$

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Re-Attempt Weekly Quiz Competition

Circles Test 1

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Circles Test 1

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SHARING IS CARING

Question 1

$$ \displaystyle \frac{\sqrt{3}}{\sqrt{2}}$$

$$ \displaystyle \frac{\sqrt{3}}{2}$$

$$ \displaystyle \frac{1}{2}$$

$$ \displaystyle \frac{1}{4}$$

Solution

Question 2

$$\mathrm{x}^{2}+\mathrm{y}^{2}-2\mathrm{x}+2\mathrm{y}-23=0$$

$$\mathrm{x}^{2}+\mathrm{y}^{2}-2\mathrm{x}-2\mathrm{y}-23=0$$

$$\mathrm{x}^{2}+\mathrm{y}^{2}+2\mathrm{x}+2\mathrm{y}-23=0$$

$$\mathrm{x}^{2}+\mathrm{y}^{2}+2\mathrm{x}-2\mathrm{y}-23=0$$

Solution

Question 3

$$x^{2}+y^{2}-12x+24=0$$

$$x^{2}+y^{2}+12x+24=0$$

$$x^{2}+y^{2}+24x-12=0$$

$$x^{2}+y^{2}-24x-12=0$$

Solution

Question 4

$$x^2 + y^2 - 4x - 4y + 16 = 0$$

$$x^2 + y^2 - 8x - 8y + 16 = 0$$

$$x^2 + y^2 + 4x + 4y + 4 = 0$$

$$x^2 + y^2 - 4x - 4y + 4 = 0$$

Solution

Question 5

5

10

6

8

Solution

Question 6

True

False

Neither

Either

Solution

Question 7

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

$$\displaystyle \left ( x-1 \right )^{2}+\left ( y+3 \right )^{2}=25$$

$$\displaystyle \left ( x-1 \right )^{2}+\left ( y-3 \right )^{2}=25$$

$$\displaystyle \left ( x+1 \right )^{2}+\left ( y-3 \right )^{2}=25$$

Solution

Question 8

$$15\pi$$

$$225\pi$$

$$20\pi$$

$$17\pi$$

Solution

Question 9

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

$$x^2-y^2=r$$

$$x^2+y^2=1$$

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

Solution

Question 10

$$\dfrac {16}9$$

$$\dfrac 43$$

$$4$$

$$\dfrac 83$$

Solution

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