Circles Test 8

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Circles Test 8

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Weekly Quiz Competition

Question 1
1 / -0

The lines $$2x - 3y = 5 \ \ \& \ \ 3x - 4y = 7$$ are diameters of a circle of area 154 sq units. Then the equation of the circle is

A
$$\displaystyle x^{2}+y^{2}+2x-2y=62 $$

B
$$\displaystyle x^{2}+y^{2}-2x+2y=47 $$

C
$$\displaystyle x^{2}+y^{2}+2x-2y=47 $$

D
$$\displaystyle x^{2}+y^{2}-2x+2y=62 $$

Solution

Intersection of any 2 diameters gives us the center

$$2x – 3y = 5 -eq.1$$

$$3x – 4y = 7 -eq.2$$

$$3 \times eq.1 – 2 \times eq.2$$

$$\implies -9y + 8y = 15 - 14$$

$$\implies y = -1 \implies x = 1$$

Center $$= (1,-1)$$

And given $$A = 154$$

$$\implies \pi r^2 = 154$$

$$r^2 = \dfrac{154}{22} \times 7 = 49$$

$$r = 7$$

Equation of circle is

$$(x - 1) ^2 + (y+1)^2 = 7^2$$

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

For the points on the circle $$\displaystyle x^{2}+y^{2}-2x-2y+1=0$$, the sum of maximum and minimum values of $$4x + 3y$$ is

A
$$\displaystyle \frac{26}{3}$$

B
$$10$$

C
$$12$$

D
$$14$$

Solution

The given equation can be written as $$(x - 1) ^2 + (y - 1)^2 = 1^2$$

Any point on this circle is given by $$(1 + \cos \theta , 1 + \sin \theta)$$

$$4x + 3y = 7 + 4 \cos \theta + 3 \sin \theta = f(\theta)$$

If $$f = a \cos \theta + b \sin \theta + c$$

$$max = c + \sqrt{a^2 + b^2} , min = c - \sqrt{a^2 + b^2}$$

$$\implies max \, f(\theta) + min \, f(\theta) = 2c =2 \times 7 = 14$$
Question 3
1 / -0

The lines $$2x - 3y = 5$$ & $$3x - 4y = 7$$ are diameters of a circle of area $$154$$ sq units Then the equation of the circle is

A
$$\displaystyle x^{2}+y^{2}+2x-2y=62 $$

B
$$\displaystyle x^{2}+y^{2}-2x+2y=47 $$

C
$$\displaystyle x^{2}+y^{2}+2x-2y=47 $$

D
$$\displaystyle x^{2}+y^{2}-2x+2y=62 $$

Solution

Intersection of any 2 diameters gives us the center

$$2x – 3y = 5 -eq.1$$

$$3x – 4y = 7 -eq.2$$

$$3 \times eq.1 – 2 \times eq.2$$

$$\implies -9y + 8y = 15 - 14$$

$$\implies y = -1 \implies x = 1$$

Center = (1,-1)

And given A = 154

$$\implies \pi r^2 = 154$$

$$r^2 = \dfrac{154}{22} \times 7 = 49$$

$$r = 7$$

Equation of circle is

$$(x - 1) ^2 + (y+1)^2 = 7^2$$

$$\implies x^2 + y^2 -2x + 2y – 47 = 0$$
Question 4
1 / -0

The centre of a circle is $$( x -2 , x+1 )$$ and it passes through the points $$( 4 , 4 )$$ Find the value ( or values ) of $$x$$, if the diameter of the circle is of length $$\displaystyle 2\sqrt{5}$$ units.

A
$$1$$ or $$3$$

B
$$-1$$ or $$4$$

C
$$5$$ or $$4$$

D
$$3$$ or $$-2$$

Solution

Radius of the circle $$ = $$ dist. between the center and given pt. on the circle.
Distance between two points $$

\left( { x }_{ 1 },{ y }_{ 1 } \right) $$ and $$ \left( { x }_{ 2 },{ y }_{ 2 }

\right) $$ can be calculated using the formula
$$ \sqrt { \left( { x }_{ 2 }-{

x }_{ 1 } \right) ^{ 2 }+\left( { y }_{ 2 }-{ y }_{ 1 } \right) ^{ 2 } } $$

Distance between the points $$ (x-2,x+1) $$ and D $$ (4,4) $$
$$= \sqrt { \left( 4-x + 2 \right) ^{ 2 }+\left( 4 - x - 1 \right) ^{ 2 } } $$
$$= \sqrt { \left( 6-x \right) ^{ 2 }+\left( 3 - x \right) ^{ 2 } } $$
$$= \sqrt { 36 + {x}^{2} - 12x + 9 + {x}^{2} - 6x } = \sqrt { 2{x}^{2} -18x + 45 } $$
Given, diameter $$ = 2 \sqrt {5} $$ $$\Rightarrow$$ Radius $$ = \sqrt {5} $$
$$ \Rightarrow \sqrt { 2{x}^{2} -18x + 45 }

= \sqrt {5} $$
Squaring both sides,
$$ 2{x}^{2} -18x + 45 = 5 $$
$$ 2{x}^{2} -18x + 40 = 0 $$
$$ {x}^{2} -9x + 20 = 0 $$
$$ (x-5)(x-4) = 0 $$
$$ x = 4 $$ or $$ 5 $$
Question 5
1 / -0

The equation circle whose center is $$(0,0)$$ and radius is $$4$$ is

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

B
$$x^2+y^2=16$$

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

D
None.

Solution

The equation of circle is $$x^2+y^2=r^2$$
Here the radius is $$4$$
So the equation is $$x^2+y^2=4^2\\x^2+y^2=16$$
Question 6
1 / -0

The equation of the circle drawn with the focus of the parabola $$(x-1)^2 - 8y = 0$$ as its centre and touching the parabola at its vertex is

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

B
$$x^2+y^2 - 4y + 1=0$$

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

D
$$x^2 + y^2 - 2x - 4y + 1=0$$

Solution

$$(x-1)^{2}=8y$$
REF.Image
$$a=2$$
focus = $$(1,2)$$ center of circle
Radius of circle = $$2$$
$$\Rightarrow $$ Eq$$^n$$ $$\Rightarrow (x-1)^{2}+(y-2)^{2}=4$$
$$\Rightarrow \boxed {x^{2}-2x+y^{2}-4y+1=0}$$ Ans
D is correct option.
Question 7
1 / -0

Find the center-radius form of the equation of the circle with center $$\left( 4,0 \right) $$ and radius $$7$$

A
$${ \left( x-4 \right) }^{ 2 }+{ y }^{ 2 }=49$$

B
$${ x }^{ 2 }+{ \left( y+4 \right) }^{ 2 }=7$$

C
$${ x }^{ 2 }+{ \left( y-4 \right) }^{ 2 }=7$$

D
$${ \left( x+4 \right) }^{ 2 }+{ y }^{ 2 }=49$$

Solution

If $$(-g,-f)$$ is the center and $$r$$ is radius

The $$(x + g) ^2 + (y+ f)^2 = r^2 $$ is the equation of the circle

There $$C = (4,0) , r = 7$$

$$\implies (x - 4) ^2 + (y - 0)^2 = 7^2$$

$$(x – 4) ^2 + (y)^2 = 49$$
Question 8
1 / -0

Two circles touch each other externally at C and a common tangent touches them at A and B. Which one is true?

A
$$CD\perp AB$$

B
$$\angle ACB=90^o$$

C
$$AC=BC$$

D
All of these

Solution

According to the question

Lets suppose that,

X and Y are two circle touch each other at P.

AB is the common tangent to circle X and Y at point A and B.

According In the given figer,

In triangle $$PAC,\ \angle CAP=\angle APC=\alpha $$

Similarly $$CB=CP,\ \angle CPB=\angle PBC=\beta $$

Now triangle APB,

$$ \angle PAB+\angle PBA+\angle APB=180 $$

$$ \alpha +\beta +\left( \alpha +\beta \right)=180 $$

$$ 2\alpha +2\beta =180 $$

$$ \alpha +\beta =90 $$

$$ \therefore \ \angle APB=90=\alpha +\beta . $$

This is the required solution.
Question 9
1 / -0

The equation to the circle with centre $$(2, 1)$$ and touching the line $$3x + 4y = 5$$ is

A
$$x^{2} + y^{2} - 4x - 2y + 5 = 0$$

B
$$x^{2} + y^{2} - 4x - 2y - 5 = 0$$

C
$$x^{2} + y^{2} - 4x - 2y + 4 = 0$$

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

Solution

The distance from the line $$ax+by+c=0$$ from the line to a point $$(x_0,y_0)$$ is:
$$D=\dfrac { \left| a{ x }_{ 0 }+b{ y }_{ 0 }+c \right| }{ \sqrt { a^{ 2 }+{ b }^{ 2 } } }$$
Here, the distance is equal to the radius of the circle and the distance from the line $$3x+4y-5=0$$ to a point $$(2,1)$$ is:
$$r=\dfrac { \left| (3\times 2)+(4\times 1)+(-5) \right| }{ \sqrt { 3^{ 2 }+{ 4 }^{ 2 } } } =\dfrac { \left| 6+4-5 \right| }{ \sqrt { 9+16 } } =\dfrac { \left| 5 \right| }{ \sqrt { 25 } } =\dfrac { 5 }{ 5 } =1$$
Therefore, equation of the circle with centre $$(2,1)$$ and radius $$1$$ is:
$${ \left( x-2 \right) }^{ 2 }+{ \left( y-1 \right) }^{ 2 }={ \left( 1 \right) }^{ 2 }\\ \Rightarrow { x }^{ 2 }+4-4x+{ y }^{ 2 }+1-2y=1\\ \Rightarrow { x }^{ 2 }+{ y }^{ 2 }-4x-2y+5=1\\ \Rightarrow { x }^{ 2 }+{ y }^{ 2 }-4x-2y+4=0$$
Hence, the equation of circle is $$x^2+y^2-4x-2y+4=0$$.
Question 10
1 / -0

What is the equation of a circle with center (-3,1) and radius 7?

A
$$\displaystyle \left ( x-3 \right )^{2}+\left ( y+1 \right )^{2}=7$$

B
$$\displaystyle \left ( x-3 \right )^{2}+\left ( y+1 \right )^{2}=49$$

C
$$\displaystyle \left ( x+3 \right )^{2}+\left ( y-1 \right )^{2}=7$$

D
$$\displaystyle \left ( x+3 \right )^{2}+\left ( y-1 \right )^{2}=49$$

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 equation of the circle is $$(x+3)^2+(y-1)^2=7^2=49$$

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Re-Attempt Weekly Quiz Competition

Circles Test 8

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Circles Test 8

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SHARING IS CARING

Question 1

$$\displaystyle x^{2}+y^{2}+2x-2y=62 $$

$$\displaystyle x^{2}+y^{2}-2x+2y=47 $$

$$\displaystyle x^{2}+y^{2}+2x-2y=47 $$

$$\displaystyle x^{2}+y^{2}-2x+2y=62 $$

Solution

Question 2

$$\displaystyle \frac{26}{3}$$

$$10$$

$$12$$

$$14$$

Solution

Question 3

$$\displaystyle x^{2}+y^{2}+2x-2y=62 $$

$$\displaystyle x^{2}+y^{2}-2x+2y=47 $$

$$\displaystyle x^{2}+y^{2}+2x-2y=47 $$

$$\displaystyle x^{2}+y^{2}-2x+2y=62 $$

Solution

Question 4

$$1$$ or $$3$$

$$-1$$ or $$4$$

$$5$$ or $$4$$

$$3$$ or $$-2$$

Solution

Question 5

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

$$x^2+y^2=16$$

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

None.

Solution

Question 6

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

$$x^2+y^2 - 4y + 1=0$$

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

$$x^2 + y^2 - 2x - 4y + 1=0$$

Solution

Question 7

$${ \left( x-4 \right) }^{ 2 }+{ y }^{ 2 }=49$$

$${ x }^{ 2 }+{ \left( y+4 \right) }^{ 2 }=7$$

$${ x }^{ 2 }+{ \left( y-4 \right) }^{ 2 }=7$$

$${ \left( x+4 \right) }^{ 2 }+{ y }^{ 2 }=49$$

Solution

Question 8

$$CD\perp AB$$

$$\angle ACB=90^o$$

$$AC=BC$$

All of these

Solution

Question 9

$$x^{2} + y^{2} - 4x - 2y + 5 = 0$$

$$x^{2} + y^{2} - 4x - 2y - 5 = 0$$

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

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

Solution

Question 10

$$\displaystyle \left ( x-3 \right )^{2}+\left ( y+1 \right )^{2}=7$$

$$\displaystyle \left ( x-3 \right )^{2}+\left ( y+1 \right )^{2}=49$$

$$\displaystyle \left ( x+3 \right )^{2}+\left ( y-1 \right )^{2}=7$$

$$\displaystyle \left ( x+3 \right )^{2}+\left ( y-1 \right )^{2}=49$$

Solution

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