Circles Test 2

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Circles Test 2

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Weekly Quiz Competition

Question 1
1 / -0

A circle has a diameter whose ends are at (-3, 2) and (12, -6) Its Equation is

A
$$\displaystyle 4x^{2}+4y^{2}-36x+16y+192=0$$

B
$$\displaystyle 4x^{2}+4y^{2}-36x+16y-192=0$$

C
$$\displaystyle 4x^{2}+4y^{2}-36x-16y-192=0$$

D
$$\displaystyle 4x^{2}+4y^{2}-36x-16y+192=0$$

Solution

Coordinate of diameter is $$(-3,2)$$ and $$(12,-6)$$
Centre of the circle is the midpoint of the diameter.
$$\therefore (x_1,y_1)=(\dfrac{-3+12}{2}, \dfrac{2-6}{2})$$
$$=(\dfrac{9}{2}, \dfrac{-4}{2})$$
$$=(\dfrac{9}{2}, -2)$$
Half the distance between $$(-3,2)$$ and $$(12,-6)$$ is the radius.
$$\therefore 2r=\sqrt {(-3-12)^2+(2+(-6))^2}$$
$$=>2r=\sqrt {15^2+8^2}$$
$$=>2r=\sqrt {225+64}$$
$$=>2r=17$$
$$=>r=\dfrac{17}{2}$$
$$\therefore Equation of circle is
$$x-(\dfrac{9}{2})^2+(y-(-2))^2=(\dfrac{17}{2})^2$$
$$=>\dfrac{(2x-9)^2}{4}+(y+2)^2=\dfrac{289}{4}$$
$$=>4x^2-36x+81+4(y^2+4y+4)-289=0$$
$$=>4x^2-36x+81+4y^2+16y+16-289=0$$
$$=>4x^2+4y^2-36x+16y+192=0$$
Question 2
1 / -0

Find the equation of a circle with center $$(0, 0)$$ and radius $$5$$.

A
$$x^2+y^2=5$$

B
$$x^2-y^2=25$$

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

D
$$(x-1)^2+(y+1)^2=25$$

Solution

Compare the equation with the standard form with center at $$(h, k)$$ and radius $$r$$ is: $$(x-h)^2+(y-k)^2=r^2$$
Substitute the value of $$(h, k) = (0, 0)$$ and $$r = 5$$
Then, the equation becomes $$x^2+y^2 = 25$$
Question 3
1 / -0

What is the radius of the circle with the following equation?
$$\displaystyle x^{2}-6x+y^{2}-4y-12=0$$

A
$$3.46$$

B
$$5$$

C
$$7$$

D
$$6$$

Solution

Equation of circle is
$$x^2-6x+y^2-4y-12=0$$
$$=>x^2-2.3.x+y^2-2.2y-12=0$$
$$=>x^2-2.3.x+3^2+y^2-2.2.y+2^2=12+3^2+2^2$$
$$=>(x-3)^2+(y-2)^2=25$$
$$=>(x-3)^2+(y-2)^2=5^2$$
$$\therefore$$ Radius of the circle=$$5$$
Question 4
1 / -0

The circle with radius $$1$$ and centre being foot of the perpendicular from $$(5, 4)$$ on y-axis, is?

A
$$x^2+y^2-8x-15=0$$

B
$$x^2+y^2-10x+24=0$$

C
$$x^2+y^2-8y+15=0$$

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

Solution

Foot of perpendicular of (5,4) on y-axis is $$(0,4)$$
Therefore the equation of circle with radius $$1cm$$ is
$${ (x-0) }^{ 2 }+{ (y-4) }^{ 2 }=1$$
$$\Rightarrow { x }^{ 2 }+{ y }^{ 2 }-8y+16=1$$
$$\Rightarrow { x }^{ 2 }+{ y }^{ 2 }-8y+15=0$$
Question 5
1 / -0

Equation of the circle with centre on y-axis and passing through the points $$(1,0),(1,1)$$ is:

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

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

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

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

Solution

Let the centre of the circle be $$(0,k)$$.
Then, the equation of the circle is $$(x-0)^2+(y-k)^2=r^2$$
Circle passes through the points $$(1,0)$$ and $$(1,1)$$.

Thus, $$1+k^2=r^2....(1)$$ and

$$1+(1-k)^2=r^2\Rightarrow 2+k^2-2k=r^2.....(2)$$

$$Eq. (1) - Eq.(2)$$ We get

$$2k-1=0$$

Solving these equations, we get, $$k=\dfrac{1}{2}$$ and $$r^2=\dfrac{5}{4}$$

Thus, the required equation of the circle is :
$$x^2+(y-\dfrac{1}{2})^2=\dfrac{5}{4}$$ i.e., $$x^2+y^2-y-1=0$$
Question 6
1 / -0

Centres of the three circles
$${x}^{2}+{y}^{2}-4x-6y-14=0$$
$${x}^{2}+{y}^{2}+2x+4y-5=0$$ and
$${x}^{2}+{y}^{2}-10x-16y+7=0$$. The centres of the circles are:

A
are the vertices of a right angle

B
the vertices of an isosceles triangle which is not regular

C
vertices of a regular triangle

D
are collinear

Solution

$${x^2} + {y^2} - 4x - 6y - 14 = 0$$ ---(i)
$${x^2} + {y^2} + 2x + 4y - 5 = 0$$ ---(ii)
$${x^2} + {y^2} - 10x - 16y + 7 = 0$$ ---(iii)
centre of the circle (i) $${c_1} = \left( { - g, - f} \right) = \left( {2,3} \right)$$
centre of the circle (ii) $${c_2} = \left( { - g, - f} \right) = \left( { - 1, - 2} \right)$$
centre of the circle (iii) $${c_3} = \left( { - g, - f} \right) = \left( {5,8} \right)$$
Now, $${c_1}{c_2} = \sqrt {{{\left( {2 + 1} \right)}^2} + {{\left( {3 + 2} \right)}^2}} = \sqrt {34} $$
$${c_2}{c_3} = \sqrt {{{\left( {5 + 1} \right)}^2} + {{\left( {8 + 2} \right)}^2}} = 2\sqrt {34} $$
$${c_1}{c_3} = \sqrt {{{\left( {5 - 2} \right)}^2} + {{\left( {8 - 3} \right)}^2}} = \sqrt {34} $$
Since $${c_1}{c_2} + {c_1}{c_3} = {c_2}{c_3}$$
Therefore $${c_1}$$,$${c_2}$$ and $${c_3}$$ collinear.
Question 7
1 / -0

Centre of circle whose normal's are $$x^{2}-2xy-3x+6y=0$$, is

A
$$\left(3,\ \dfrac{3}{2}\right)$$

B
$$\left(3,\ -\dfrac{3}{2}\right)$$

C
$$\left(\dfrac{3}{2},\ 3\right)$$

D
$$None\ of\ these$$

Solution

$${x}^{2} - 2xy - 3x + 6y = 0$$

$$\Rightarrow \; \left( x - 3 \right) \left( x - 2y \right) = 0$$

Hence, $$x = 3$$ and $$x = 2y$$ are two normals.

The intersection point of these two normals will be the centre of the circle.
$$\therefore$$ for $$x = 3$$,

$$\Rightarrow$$$$y = \cfrac{x}{2} = \cfrac{3}{2}$$

Hence, the intersection point is $$\left( 3, \cfrac{3}{2} \right)$$.
Hence, the centre of the given circle is $$\left( 3, \cfrac{3}{2} \right)$$.
Question 8
1 / -0

The equation of the circle passing through $$(3, 6)$$ and whose centre is $$(2, -1)$$ is

A
$${x^2} + {y^2} - 4x + 2y = 45$$

B
$${x^2} + {y^2} - 2y + 45 = 0$$

C
$${x^2} + {y^2} + 4x - 2y = 45$$

D
$${x^2} + {y^2} + 2y + 45 = 0$$

Solution

Let the equation be $$x^2 + y^2 + fx + gy + d = 0$$ with centre =$$(-\cfrac{-f}{2} , -\cfrac{g}{2})$$
Hence, $$f = -4 , g = 2$$
Equation becomes = $$x^2+ y^2 - 4x +2y+d = 0$$
Putting x = 3 y = 6 in the above equation we get
$$9+36-12+12+d = 0$$
$$\implies d = -45$$
$$equation = x^2+y^2-4x+2y-45 = 0$$
Question 9
1 / -0

The radius of the circle centred at $$(4,5)$$ and passing through the centre of the circle $${x}^{2}+{y}^{2}+4x+6y-12=0$$ is

A
$$2\sqrt {5}$$

B
$$2\sqrt {10}$$

C
$$3\sqrt {5}$$

D
$$3\sqrt {10}$$

Solution

Let the center $$A(4,5)$$
$$x^2+y^2+4x+6y-12=0$$
On comparing with it
$$x^2+y^2+2gx+2fy+c=0$$
$$g=2,f=3,c=-12$$
center of the circle $$=(-g,-f)=(-2,-3)$$
Then required radius of the circle \
$$=\sqrt{(4+2)^2+(5+3)^2}=10$$
Question 10
1 / -0

The centre of the circle $$x^2+y^2+10x-20y+100=0$$ is

A
$$(5,10)$$

B
$$(-5,10)$$

C
$$(-5,-10)$$

D
$$(5,-10)$$

Solution

Given the equation of the circle is
$$x^2+y^2+10x-20y+100=0$$
or, $$x^2+10x+25+y^2-20y+100=25$$
or, $$(x+5)^2+(y-10)^2=5^2$$
From this equation it is clear that the centre is $$(-5,10)$$.

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Circles Test 2

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Circles Test 2

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Question 1

$$\displaystyle 4x^{2}+4y^{2}-36x+16y+192=0$$

$$\displaystyle 4x^{2}+4y^{2}-36x+16y-192=0$$

$$\displaystyle 4x^{2}+4y^{2}-36x-16y-192=0$$

$$\displaystyle 4x^{2}+4y^{2}-36x-16y+192=0$$

Solution

Question 2

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

$$x^2-y^2=25$$

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

$$(x-1)^2+(y+1)^2=25$$

Solution

Question 3

$$3.46$$

$$5$$

$$7$$

$$6$$

Solution

Question 4

$$x^2+y^2-8x-15=0$$

$$x^2+y^2-10x+24=0$$

$$x^2+y^2-8y+15=0$$

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

Solution

Question 5

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

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

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

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

Solution

Question 6

are the vertices of a right angle

the vertices of an isosceles triangle which is not regular

vertices of a regular triangle

are collinear

Solution

Question 7

$$\left(3,\ \dfrac{3}{2}\right)$$

$$\left(3,\ -\dfrac{3}{2}\right)$$

$$\left(\dfrac{3}{2},\ 3\right)$$

$$None\ of\ these$$

Solution

Question 8

$${x^2} + {y^2} - 4x + 2y = 45$$

$${x^2} + {y^2} - 2y + 45 = 0$$

$${x^2} + {y^2} + 4x - 2y = 45$$

$${x^2} + {y^2} + 2y + 45 = 0$$

Solution

Question 9

$$2\sqrt {5}$$

$$2\sqrt {10}$$

$$3\sqrt {5}$$

$$3\sqrt {10}$$

Solution

Question 10

$$(5,10)$$

$$(-5,10)$$

$$(-5,-10)$$

$$(5,-10)$$

Solution

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