Circles Test 4

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Circles Test 4

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Weekly Quiz Competition

Question 1
1 / -0

Which is not represented by quadratic equation ?

A
Circle

B
Straight line

C
Parabola

D
Hyperbola

Solution

Equation of circle is given by:
$$x^2+y^2+2gx+2fy+c$$
Equation of straight line is given by:
$$y=mx+c$$
Equation of parabola is given by:
$$(y-k)^2=4p(x-h)^2$$
Equation of hyperbola is given by:
$$\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$$
Hence, we can see from the above equations that only equation of a straight line is not represented by quadratic equation.
Question 2
1 / -0

Find the area of $$x^2+y^2=49$$

A
154

B
49

C
88

D
None

Solution

Th eequation $$x^2+y^2=49$$ describes a circle with $$7$$ as radius
So the area is given as $$\pi r^2\\\dfrac{22}{7}\times 7^2=154$$
Question 3
1 / -0

The equation $${ x }^{ 2 }+{ y }^{ 2 }=9$$ meets x-axis at

A
$$(\pm 3,0)$$

B
$$(\pm 9,0)$$

C
$$(\pm 1,0)$$

D
(-5/14, 5/14)

Solution

$$x^2+y^2=9$$
The equation meets x-axis if the y-coordinate is $$0$$
So $$x^2+0=9\\x^2=9\\x=\sqrt9\\x=\pm 3$$
So the point is $$(\pm 3,0)$$
Question 4
1 / -0

Find the Center of circle $$x^2+y^2-4x-8x+25=0$$

A
$$(2,4)$$

B
$$(-2,-4)$$

C
$$(4,2)$$

D
$$(-4,-2)$$

Solution

The general equation of center of circle $$x^2+y^2+2gx+2fy+c=0$$ is $$(-g,-f)$$

So the center of circle $$x^2+y^2-4x,-8x+25=0$$ is $$(2,4)$$
Question 5
1 / -0

Radius of the circle $$2x^2+2y^2+8x+4y-3=0$$ is

A
$$\sqrt{17}$$

B
$$\sqrt{23}$$

C
$$\sqrt{\dfrac{13}{2}}$$

D
$$\sqrt{\dfrac{11}{2}}$$

Solution

Formula,

$$r=\sqrt{g^2+f^2-c}$$

From given, we have,

$$2g=8\Rightarrow g=4$$

$$2f=4\Rightarrow f=2$$

$$c=-3$$

$$r=\sqrt{4^2+2^2-(-3)}=\sqrt{23}$$
Question 6
1 / -0

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

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

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

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

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

Solution

Equation of circle with centre (a,b) and radius r,
$$\rightarrow (x-a)^2 + (y-b)^2 = r^2$$

For the circle to have minimum radius, given points must be diametric points,
So, Centre of circle = $$(\dfrac{2+0}{2},\dfrac{0+4}{2}) = (1,2)$$
Radius of circle = $$\sqrt{(1-0)^2 + (2-4)^2} = \sqrt{5}$$

So, The equation of circle is $$(x-1)^2+(y-2)^2 = 5$$
$$\rightarrow x^2+y^2-2x-4y=0$$
Question 7
1 / -0

The locus of a point which moves such that the sum of the squares of its distances from three vertices of a triangle ABC is constant is a circle
whose centre is at the

A
centroid of triangle ABC

B
Circumcentre of triangle ABC

C
Orthocentre of triangle ABC

D
incentre of triangle ABC

Solution

$$AP^{2}+AB^{2}+PC^{2}=C$$
$$(x-x_{1})^{2}+(y-y_{1})^{2}+(x-x_{2})^{2}-(y-y_{2})^{2}+(x-x_{3})^{2}+(y-y_{3})^{2}=c$$
$$3x^{2}-(2x_{1}+2x_{2}+2x_{3})x+x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+3y^{2}-2(y_{1}+y_{2}+y_{3})+y_{1}^{2}+y_{2}^{2}+y_{3}^{2}=c$$
$$x^{2}+y^{2}-\frac{2}{3}(x_{1}+x_{2}+x_{3})x-\frac{2}{3}(y_{1}+y_{2}+y_{3})c+(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+y_{1}^{2}+y_{2}^{2}+y_{3}^{2}-c)=0$$
So, center of this circle is $$(\dfrac{x_{1}+x_{2}+x_{3}}{3},\dfrac{y_{1}+y_{2}+y_{3}}{3}),$$
centroid of triangle is a center of that circle.
Question 8
1 / -0

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

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

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

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

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

Solution

$$2x-3y=5----(1)$$
$$3x-4y=7----(2)$$
are diameter of circle so, point of intersection is centre of circle.
from (1) & (2), $$x=1, y=-1$$
$$\therefore $$ area of circle is $$154$$
$$=\pi r^{2}=154$$
$$r^{2}=\dfrac{7}{22}\times154$$
$$=\dfrac{7\times14}{2}$$
$$r^{2}=49$$
$$r=7$$
$$\therefore$$ Equation of circle with centre $$(1,-1)$$ and radius $$7$$ units
$$(x-1)^{2}+(y+1)^{2}=7^{2}$$
$$\Rightarrow x^{2}+y^{2}-2x+2y-47=0$$
Question 9
1 / -0

lf the line $$3{x}-2{y}+6=0$$ meets $${x}$$-axis, $$y$$-axis respectively at $${A}$$ and $${B}$$, then the equation of the circle with radius $${A}{B}$$ and Centre at $${A}$$ is

A
$$x^{2}+y^{2}+4x+9=0$$

B
$$x^{2}+y^{2}+4x-9=0$$

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

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

Solution

Given that:
Equation of the straight line: $$3x-2y+6=0$$
Since, $$A$$ is the centre and $$B$$ is one end-point.
So, $$A$$ is the mid-point of two diameter end points.
One point is $$B$$ and let other be $$(x,y)$$
So, $$-2=\dfrac{x+0}{2}, 0=\dfrac{y+3}{2}$$
So,Co-ordinates of other end-point$$=\left(-4,-3\right)$$
Equation of the circle as centre $$A$$ is
$$(x-0)(x+4)+(y-3)(y+3)=0$$ (Diameter form)
or, $$x^2+4x+y^2-9=0$$
or, $$x^2+y^2+4x-9=0$$
Hence, B is the correct option.
Question 10
1 / -0

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

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

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

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

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

Solution

$$2x+3y+1=0$$ and $$3x-y-4=0$$ lie along diameter of circle.

So, point of intersection of $$2x+3y+1=0$$ and $$3x-7-4=0$$ is a centre of required circle.

$$\therefore 11x-11=0.$$

$$x=1$$ & $$y=-1.$$

So, equation of circle having radius $$r=\dfrac{10}{2\pi}=5$$ and centre $$(1,-1)$$ is given by $$(x-1)^{2}+(y+1)^{2}=25=r^{2}$$

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

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Re-Attempt Weekly Quiz Competition

Circles Test 4

Result

Circles Test 4

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Rank

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SHARING IS CARING

Question 1

Circle

Straight line

Parabola

Hyperbola

Solution

Question 2

154

49

88

None

Solution

Question 3

$$(\pm 3,0)$$

$$(\pm 9,0)$$

$$(\pm 1,0)$$

(-5/14, 5/14)

Solution

Question 4

$$(2,4)$$

$$(-2,-4)$$

$$(4,2)$$

$$(-4,-2)$$

Solution

Question 5

$$\sqrt{17}$$

$$\sqrt{23}$$

$$\sqrt{\dfrac{13}{2}}$$

$$\sqrt{\dfrac{11}{2}}$$

Solution

Question 6

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

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

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

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

Solution

Question 7

centroid of triangle ABC

Circumcentre of triangle ABC

Orthocentre of triangle ABC

incentre of triangle ABC

Solution

Question 8

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

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

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

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

Solution

Question 9

$$x^{2}+y^{2}+4x+9=0$$

$$x^{2}+y^{2}+4x-9=0$$

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

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

Solution

Question 10

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

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

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

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

Solution

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