Circles Test 9

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Circles Test 9

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Weekly Quiz Competition

Question 1
1 / -0

An ambulance company provides services within an 80 mile radius of their headquarters If this service area is represented graphically with the headquarters located at the coordinates (0, 0) what is the equation that represents the service area?

A
$$\displaystyle x^{2}+y^{2}=80$$

B
$$\displaystyle \left ( x-0 \right )^{2}+\left ( y-0 \right )^{2}=80$$

C
$$\displaystyle x^{2}+y^{2}=1600$$

D
$$\displaystyle x^{2}+y^{2}=6400$$

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
$$x^2+y^2=80^2=6400$$ option D
Question 2
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The radius of the circle $$x^{2} + y^{2} + 4x + 6y + 13 = 0$$ is

A
$$\sqrt {26}$$

B
$$\sqrt {13}$$

C
$$\sqrt {23}$$

D
$$0$$

Solution

Given equation is
$$x^{2} + y^{2} + 4x + 6y + 13 = 0$$
or $$(x^{2} + 4x + 4) + (y^{2} + 6y + 9) + 13 = 4 + 9$$
or $$(x + 2)^{2} (y + 3)^{2} = 0$$
$$\therefore$$ Radius of circle $$= 0$$.
Question 3
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Graph the circle
$${ \left( x-3 \right) }^{ 2 }+{ \left( y+4 \right) }^{ 2 }=4$$

A

B

C

D

Solution

$$(x - 3) ^2 + (y+ 4)^2 = 4$$

$$\implies center = (3,-4) , r = 2$$

Since $$C = (3,-4)$$
C lies in the fourth quadrant with radius $$2$$
Question 4
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The equation of a circle which has a tangent $$3x+4y=6$$ and two normals given by $$(x-1)(y-2)=0$$ is

A
$$(x-3)^2+(y-4)^2=5^2$$

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

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

D
$$x^2+y^2-2x-4y+5=0$$

Solution

Equation of tangent $$=3x+4y=6$$ and two normals are $$(x-1)(y-2)=0$$
$$\Rightarrow x-1=0\,$$ and $$\, y-2=0$$
$$\therefore \text{Radius}=\displaystyle \frac{3(1)+4(2)-6}{\sqrt{9+16}}$$
$$=\displaystyle \frac{5}{5}=1$$
$$\therefore$$ Equation of the circle is
$$(x-1)^2+(y-2)^2=1\, $$ or$$ \, x^2+y^2-2x-4y+4=0$$
Question 5
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Find the equation of the circle with center on x + y = 4 and 5x + 2y + 1 = 0 and having a radius of 3

A
$$\displaystyle x^{2}+y^{2}+6x-16y+64=0$$

B
$$\displaystyle x^{2}+y^{2}+8x-14y+25=0$$

C
$$\displaystyle x^{2}+y^{2}+6x-14y+49=0$$

D
$$\displaystyle x^{2}+y^{2}+6x-14y+36=0$$

E
none of these

Solution

Given,
Centre of circle is on $$x+y=4$$ and $$5x+2y+1=0$$
Thus,
$$x+y=4$$
$$=>y=4-x$$ (i)
$$5x+2y+1=0$$ (ii)
$$=>5x+2(4-x)+1=0$$
$$=>5x+8-2x+1=0$$
$$=>3x+9=0$$
$$=>x=-3$$
$$\therefore y=4-(-3)$$
$$=7$$
$$\therefore$$ Centre of circle is $$(-3,7)$$
Radius of circle=$$3$$ (given)
$$\therefore $$ Equation of circle is
$$(x-(-3))^2+(y-7)^2=3^2$$
$$=>(x+3)^2+(y-7)^2=9$$
$$=>(x^2+6x+9)+(y^2-14y+49)-9=0$$
$$=>x^2+6x+y^2-14y+49=0$$
Question 6
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Find the equation of k for which the equation $$\displaystyle x^{2}+y^{2}+4x-2y-k=0$$ represents a point circle

A
$$5$$

B
$$-5$$

C
$$6$$

D
$$-6$$

E
none of these

Solution

Standard form of circle is
$$(x-x_1)^2+(y-y_1)^2=r^2$$
When $$(x_1,y_1)$$ is the centre of the circle and $$r$$ is the radius of the circle.
Now,given equation is,
$$x^2+y^2+4x-2y-k=0$$
$$=>x^2+4x+y^2-2y=k$$
$$=>x^2+2.x.2+2^2+y^2-2.y.1+1^1=k+2^2+1^1$$
$$=>(x+2)^2+(y-1)^2=k+4+1$$
$$=>(x+2)^2+(y-1)^2=k+5$$
Since it represents a point circle
$$\therefore k+5=0$$
$$=>k=-5$$
Question 7
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The radius of the circle passing through the point $$(6, 2)$$ and two of whose diameters are $$\displaystyle x+y=6$$ and $$\displaystyle x+2y=4$$ is:

A
$$4$$

B
$$6$$

C
$$20$$

D
$$\displaystyle \sqrt { 20 } $$

Solution

Point of intersection of the given diameters is $$(8,-2)$$ which is the centre of the circle
Also the circle pass through the point $$(6,2)$$, so the radius is $$=\sqrt{(8-6)^2+(-2-2)^2}=\sqrt{20}$$
Question 8
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The equation of the circle whose centre and radius are $$\left( 1,-1 \right) $$ and $$4$$ respectively, is

A
$${ x }^{ 2 }-{ y }^{ 2 }+2x+2y-14=0$$

B
$${ x }^{ 2 }+{ y }^{ 2 }-2x-2y-14=0$$

C
$${ x }^{ 2 }+{ y }^{ 2 }-2x+2y-14=0$$

D
$${ x }^{ 2 }+{ y }^{ 2 }-2x-2y+14=0$$

Solution

We know that equation of circle is
$${ \left( x-h \right) }^{ 2 }+{ \left( y-k \right) }^{ 2 }={ r }^{ 2 }$$
Here, centre is $$\left( 1,-1 \right) $$ and radius is $$4$$.
$$\therefore $$ Equation is
$${ \left( x-1 \right) }^{ 2 }+{ \left( y+1 \right) }^{ 2 }={ \left( 4 \right) }^{ 2 }$$
$$\Rightarrow { x }^{ 2 }+1-2x+{ y }^{ 2 }+1+2y=16$$
$$\Rightarrow { x }^{ 2 }+{ y }^{ 2 }-2x+2y-14=0$$
Question 9
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Write the equation of the circle with center at $$(0,0)$$ and a radius of $$6$$

A
$$(x-6)^2+y^2=36$$

B
$$x^2+(y-6)^2=0$$

C
$$x^2+y^2=36$$

D
$$x^2+y^2=-36$$

Solution

An equation of the circle with center $$(h,k)$$ and radius $$r$$ is
$$(x-h)^{ 2 }+(y-k)^{ 2 }=r^{ 2 }$$
So, if the center is $$(0,0)$$ and the radius is $$6$$, an equation of the circle is: $$(x-0)^{ 2 }+(y-0)^{ 2 }=6^{ 2 }$$
On simplifying, we get:
$$x^{ 2 }+y^{ 2 }=36$$
Hence, the equation of the circle is $$x^{ 2 }+y^{ 2 }=36$$.
Question 10
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Find the equation of a circle with center $$(2,0)$$ and passing through point $$\left( 3,\sqrt { 3 } \right) $$.

A
$$(x - 2)^{2} + y^{2} = 4$$

B
$$(x + 2)^{2} + y^{2} = 4$$

C
$$(x - 2)^{2} + y^{2} = 16$$

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

Solution

Let the radius be '$$r$$'.
Then the equation of the circle will be
$$(x-2)^{2}+y^{2}=r^{2}$$
Now, substituting the point $$(3,\sqrt{3})$$ in the above equation gives us $$(3-2)^{2}+(\sqrt{3})^{2}=r^{2}$$
$$\Rightarrow 1+3^{2}=r^{2}$$ or $$r^{2}=4$$
Hence, the equation of the circle is $$(x-2)^{2}+y^{2}=4$$.