Circles Test 12

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Circles Test 12

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Weekly Quiz Competition

Question 1
1 / -0

A circle of radius $$\sqrt 8$$ is passing through origin and the point $$(4, 0)$$. If the centre lies on the line $$y = x$$, then the equation of the circle is

A
$$(x - 2)^2 + (y - 2)^2 = 8$$

B
$$(x + 2)^2 + (y + 2)^2 = 8$$

C
$$(x - 3)^2 + (y - 3)^2 = 8$$

D
$$(x + 3)^2 + (y + 3)^2 = 8$$

E
$$(x - 4)^2 + (y - 4)^2 = 8$$

Solution

Let the equation of the circle be $$(x-a)^2 + (y-a)^2 = 8 \quad [\because \,\, \text{Centre lies on } x =y \,\, \text{and radius is } \sqrt{8}]$$

Both $$(0,0)$$ and $$(4,0)$$ lies of the circle, therefore we have the below 2 equations

$$a^2+a^2 = 8 \implies a= \pm 2$$

$$(4-a)^2+a^2 = 8 \implies a^2-4a +4 = 0 \implies (a-2)^2 = 0 \implies a = 2$$

$$\therefore, \,\, a = 2$$ and the required equation is $$(x-2)^2 +(y-2)^2= 8$$
Question 2
1 / -0

The equation of the circle having $$x-y-2=0$$ and $$x-y+2=0$$ as two tangents and $$x-y=0$$ as diameter is

A
$${ x }^{ 2 }+{ y }^{ 2 }+2x-2y+1=0$$

B
$${ x }^{ 2 }+{ y }^{ 2 }-2x+2y-1=0$$

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

D
$${ x }^{ 2 }+{ y }^{ 2 }=1$$

Solution

Since, the equations of tangents $$x-y-2=0$$ and $$x-y+2=0$$ are parallel.
Therefore, distance between them $$=$$ Diameter of the circle
$$=\dfrac { \left| 2-\left( -2 \right) \right| }{ \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } }$$ $$\left( \because \dfrac { { C }_{ 2 }-{ C }_{ 1 } }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \right) $$
$$=\dfrac { 4 }{ \sqrt { 2 } } =2\sqrt { 2 } $$
Hence, radius $$=\dfrac { 1 }{ 2 } \left( 2\sqrt { 2 } \right) =\sqrt { 2 } $$
It is clear from the figure that centre lies on the origin.
Therefore, equation of circle is $${ \left( x-0 \right) }^{ 2 }+{ \left( y-0 \right) }^{ 2 }={ \left( \sqrt { 2 } \right) }^{ 2 }$$.
$$\Rightarrow { x }^{ 2 }+{ y }^{ 2 }=2$$
Question 3
1 / -0

The lines $$y=x+\sqrt { 2 } $$ and $$y=x-2\sqrt { 2 } $$ are the tangent of certain circle. If the point $$\left( 0,\sqrt { 2 } \right) $$ lies on this circle, then its equation is

A
$${ \left( x-\cfrac { 3 }{ 2\sqrt { 2 } } \right) }^{ 2 }+{ \left( y+\cfrac { 1 }{ 2\sqrt { 2 } } \right) }^{ 2 }=\cfrac { 9 }{ 4 } $$

B
$${ \left( x+\cfrac { 3 }{ 2\sqrt { 2 } } \right) }^{ 2 }+{ \left( y+\cfrac { 1 }{ 2\sqrt { 2 } } \right) }^{ 2 }=\cfrac { 9 }{ 4 } $$

C
$${ \left( x-\cfrac { 3 }{ 2\sqrt { 2 } } \right) }^{ 2 }+{ \left( y-\cfrac { 1 }{ 2\sqrt { 2 } } \right) }^{ 2 }=\cfrac { 9 }{ 4 } $$

D
None of the above

Solution

Both tangents are parallel to each other, Therefore the distance between these two lines is the diameter of the circle.
diameter $$=\dfrac{3\sqrt{2}}{\sqrt{2}}=3$$
Radius is =$$\dfrac{3}{2}$$
If $$(x_1, y_1)$$ is the centre of the circle, Then

$$\dfrac{x_1-0}{1}=\dfrac{y_1-\sqrt{2}}{-1}=\dfrac{1}{2}\dfrac{|0-\sqrt{2}-2\sqrt{2}|}{2}$$

Centre is $$\bigl(\dfrac{3}{2\sqrt{2}}, \dfrac{1}{2\sqrt{2}}\Bigr)$$
Equation of the circle is $$(x-\frac{1}{3\sqrt{2}})^2+(y-\frac{1}{2\sqrt{2}})^2=9/4$$
Question 4
1 / -0

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

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

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

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

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

Solution

Given equation of lines is $$2x-3y=5$$ ..... (i) and $$3x-4y=7$$ .....(ii)
Solving above equations, we get the point of intersection as $$(1,-1)$$.
Eqns (i) and (ii) are diameters of the circle.
We know that the centre of circle $$=$$ point of intersection of diameters $$=(1,-1)$$.
Now, it is given that the area of the circle $$=154$$.
$$\pi {r}^{2}=154$$
$$\Rightarrow r=7$$
Hence, the equation of required circle is
$${(x-1)}^{2}+{(y+1)}^{2}={7}^{2}$$
$$\Rightarrow$$ $${x}^{2}+{y}^{2}-2x+2y=47$$
Question 5
1 / -0

The equation of the circle which touches the lines $$x = 0, y = 0$$ and $$4x + 3y = 12$$ is

A
$$x^{2} + y^{2} - 2x - 2y - 1 = 0$$

B
$$x^{2} + y^{2} - 2x - 2y + 3 = 0$$

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

D
$$x^{2} + y^{2} - 2x - 2y + 1 = 0$$

E
$$x^{2} + y^{2} - 2x - 2y - 3 = 0$$

Solution

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

Since, circle touches the lines $$x = 0, y = 0$$, i.e.,

$$x$$-axis and $$y$$-axis, then its centre is $$C\equiv (r, r)$$

Now, $$\text{radius} (PC) =$$ Perpendicular distance from $$(C)$$ to the line

$$4x + 3y = 12$$

i.e., $$r = \dfrac {|4r + 3r - 12|}{\sqrt {16 + 9}}$$

$$\Rightarrow 7r - 12 = \pm 5r$$

So, $$r = 1$$ or $$6$$

But $$r\neq 6$$

Required equation of circle is

$$(x - 1)^{2} + (y - 1)^{2} = 1$$

$$\Rightarrow x^{2} + 1 - 2x + y^{2} + 1 - 2y = 1$$

$$\Rightarrow x^{2} + y^{2} - 2x - 2y + 1 = 0$$
Question 6
1 / -0

The radius of the circle passing through the points $$(2,3),(2,7)$$ and $$(5,3)$$ is

A
$$5$$

B
$$4$$

C
$$\cfrac { 5 }{ 2 } $$

D
$$2$$

E
$$\sqrt { 5 } $$

Solution

Let the given points $$A(2,3),B(5,3)$$ and $$C(2,7)$$
Diameters of circle is distance between $$B(5,3)$$ and $$C(2,7)$$
$$d=\sqrt { { (2-5) }^{ 2 }+{ (7-3) }^{ 2 } } =\sqrt { 9+16 } $$
$$d=\sqrt { 25 } =5$$
$$\therefore$$ Radius $$=\cfrac { d }{ 2 } =\cfrac { 5 }{ 2 } $$
Question 7
1 / -0

The equation of the circumcircle of the triangle formed by the lines $$y + \sqrt{3} x = 6, y - \sqrt{3} x = 6$$ and $$y=0$$ is

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

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

C
$$x^2+y^2-4y=12$$

D
$$x^2+y^2+4x=12$$

Solution

Triangle $$ABC$$ is an equilateral triangle
Hence,
$$BC=4\sqrt{3} \\ 2R=\cfrac{a}{\sin A} \\ 2R=\cfrac{4\sqrt{3}}{\sin 60^o}=8 \\ R=4$$
Circumcentre of $$ABC$$ $$\left( \cfrac{x_1+x_2+x_3}{3},\cfrac{y_1+y_2+y_3}{3} \right) \\ =(0,2) $$
Equation of circumcircle
$$(x-0)^2+(y-2)^2=4^2 \\ x^2+y^2-4y=12$$
Question 8
1 / -0

The equation of the circle whose two diameters are the lines $$x+y=4$$ and $$x-y=2$$ and which passes through $$(4 , 6 )$$ is

A
$$x^2+y^2-6x-2y-16=0$$

B
$$x^2+y^2-6x-2y=15$$

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

D
$$5(x^2+y^2)-4x=16$$

Solution

Two diameters are the lines,
$$x+y=4$$ ----- ( 1 )
$$x-y=2$$ ----- ( 2 )

So, first, we find the intersection point of diameters that is the center of the circle.
Adding equation ( 1 ) and ( 2 ) we get,
$$\Rightarrow$$ $$2x=6$$
$$\Rightarrow$$ $$x=3$$

Substituting value of $$x$$ in equation ( 1 ),
$$\Rightarrow$$ $$3+y=4$$
$$\Rightarrow$$ $$y=1$$

So, the center of circle is $$(3,1).$$
Now, circle is pass through $$(4,6)$$
The equation of circle is $$(x-a)^2+(y-b)=r^2$$, where $$(a,b)$$ co-ordinates of circle and $$r$$ is radius.

Put center coordinates and pass through point
$$\Rightarrow$$ $$(4-3)^2+(6-1)^2=r^2$$
$$\Rightarrow$$ $$1+25=r^2$$
$$\Rightarrow$$ $$r^2=26$$

So, the equation of circle is,
$$\Rightarrow$$ $$(x-3)^2+(y-1)^2=26$$
$$\Rightarrow$$ $$x^2-6x+9+y^2-2y+1=26$$
$$\Rightarrow$$ $$x^2+y^2-6x-2y-16=0$$
Question 9
1 / -0

If the lines $$3x - 4y - 7 = 0$$ and $$2s - 3y - 5 = 0$$ are two diameters of a circle of area $$49\pi$$ square units, the equation of the circle is-

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

B
$$x^{2} + y^{2} - 2x + 2y - 62 = 0$$

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

D
$$x^{2} + y^{2} + 2x - 2y - 47 = 0$$

Solution

Given: Area of circle=$$49\pi \ sq.units$$

$$\therefore$$ Area of circle=$$\pi \ r^2= 49\pi \ sq.units$$
where, $$r$$ is the radius of the circle
$$\pi \ r^2= 49\pi$$
$$r^2= 49$$
$$r= 7 \ units$$

Given eqs of lines are, $$3x-4y-7=0$$ and $$2x-3y-5=0$$

Solving the equations, we get $$x=1 \ , \ y=-1$$

The points of intersection of the lines is the center of the circle i.e., $$(1,-1)$$

We have the eqn for the circle with center $$(a,b)$$ and radius $$r$$,

$$(x-a)^2+(y-b)^2=(r)^2$$
$$(x-1)^2+(y+1)^2=(7)^2$$
$$(x^2+1-2x)+(y^2+1+2y)=49$$
$$x^2+y^2+2-2x+2y=49$$
$$x^2+y^2-2x+2y=49-2$$

$$\therefore$$ The equation of the given circle is,

$$x^2+y^2-2x+2y-47=0$$
Question 10
1 / -0

The equation of the circle passing through $$(2,0)$$ and $$(0,4)$$ and having the minimum radius is

A
$$x^{2}+y^{2}=20$$

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

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

D
$$x^{2}+y^{2}=16$$

Solution

Given points are $$(2,0)$$ and $$(0,4)$$

Therefore, equation of circle is $$(x-2)(x-0)+(y-0)(y-4)=0$$

By expanding, we get

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

Option B is correct.

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Circles Test 12

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Circles Test 12

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Question 1

$$(x - 2)^2 + (y - 2)^2 = 8$$

$$(x + 2)^2 + (y + 2)^2 = 8$$

$$(x - 3)^2 + (y - 3)^2 = 8$$

$$(x + 3)^2 + (y + 3)^2 = 8$$

$$(x - 4)^2 + (y - 4)^2 = 8$$

Solution

Question 2

$${ x }^{ 2 }+{ y }^{ 2 }+2x-2y+1=0$$

$${ x }^{ 2 }+{ y }^{ 2 }-2x+2y-1=0$$

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

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

Solution

Question 3

$${ \left( x-\cfrac { 3 }{ 2\sqrt { 2 } } \right) }^{ 2 }+{ \left( y+\cfrac { 1 }{ 2\sqrt { 2 } } \right) }^{ 2 }=\cfrac { 9 }{ 4 } $$

$${ \left( x+\cfrac { 3 }{ 2\sqrt { 2 } } \right) }^{ 2 }+{ \left( y+\cfrac { 1 }{ 2\sqrt { 2 } } \right) }^{ 2 }=\cfrac { 9 }{ 4 } $$

$${ \left( x-\cfrac { 3 }{ 2\sqrt { 2 } } \right) }^{ 2 }+{ \left( y-\cfrac { 1 }{ 2\sqrt { 2 } } \right) }^{ 2 }=\cfrac { 9 }{ 4 } $$

None of the above

Solution

Question 4

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

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

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

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

Solution

Question 5

$$x^{2} + y^{2} - 2x - 2y - 1 = 0$$

$$x^{2} + y^{2} - 2x - 2y + 3 = 0$$

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

$$x^{2} + y^{2} - 2x - 2y + 1 = 0$$

$$x^{2} + y^{2} - 2x - 2y - 3 = 0$$

Solution

Question 6

$$5$$

$$4$$

$$\cfrac { 5 }{ 2 } $$

$$2$$

$$\sqrt { 5 } $$

Solution

Question 7

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

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

$$x^2+y^2-4y=12$$

$$x^2+y^2+4x=12$$

Solution

Question 8

$$x^2+y^2-6x-2y-16=0$$

$$x^2+y^2-6x-2y=15$$

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

$$5(x^2+y^2)-4x=16$$

Solution

Question 9

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

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

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

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

Solution

Question 10

$$x^{2}+y^{2}=20$$

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

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

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

Solution

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