Circles Test 18

Result

Circles Test 18

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The equation of circle with centre $$(1,2)$$ and tangent $$x+y-5=0$$ is

A
$${x}^{2}+{y}^{2}+2x-4y+6=0$$

B
$${x}^{2}+{y}^{2}-2x-4y+3=0$$

C
$${x}^{2}+{y}^{2}-2x+4y+8=0$$

D
$${x}^{2}+{y}^{2}-2x-4y+8=0$$

Solution
Question 2
1 / -0

The circle $${x^2} + {y^2} - 3x - 4y + 2 = 0$$ cuts $$x$$-axis

A
$$(2,0),(-3,0)$$

B
$$(3,0),(4,0)$$

C
$$(1,0),(-1,0)$$

D
$$(1,0),(2,0)$$

Solution

$$x^{2}+y^{2}-3x-4y+2=0$$

x-axis will be cut when $$y=0 $$

put $$y=0 $$

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

$$(x-2)(x-1)=0$$

$$x=1,2 $$

points $$(1,0),(2,0)$$
D is correct.
Question 3
1 / -0

Equation of the circle which passes through the centre of the circle $$x^{2} + y^{2} + 8x + 10y - 7 = 0$$ and is concentric with the circle $$2x^{2} + 2y^{2} - 8x - 12y - 9 = 10$$ is

A
$$x^{2} + y^{2} - 4x - 8y - 97 = 0$$

B
$$x^{2} + y^{2} - 4x - 6y - 87 = 0$$

C
$$x^{2} + y^{2} - 2x - 8y - 95 = 0$$

D
None of these

Solution

First equation $$=x^2+y^2+8x+10y-7=0$$

its centre $$=(-4, -5)$$

Second equation of the circle

$$=2x^2+2y^2+8x-12y-9=0$$

its, centre $$=(2, 3)$$

radius:$$=r^2\sqrt {(2+4)^2+(3+5)^2}$$

$$r^2=\sqrt {6^2+8^2}=\sqrt {36+64}$$

$$r^2 =100$$

$$r=10$$

The equation of the concentric

circle $$=(x-2)^2+(y-3)^2=10^2$$

$$x^2-4x+4+y^2-6y+9-100=0$$

$$x^2-4x+y^2-6y+13-100=0$$

$$x^2+y^2-4x-6y-87=0$$
Question 4
1 / -0

Equation of the circle of radius 5 whose centre lies on y-axis in first quadrant and passes through$$\left( {3,\,\,\,\,2} \right)$$ is

A
$${x^2} + {y^2} - 12y + 11 = 0$$

B
$${x^2} + {y^2} - 6y - 1 = 0$$

C
$${x^2} + {y^2} - 8y + 3 = 0$$

D
None of these

Solution

Let the coordinate of the center be $$(0,k)$$ as center lies on y-axis

hence, the Equation of circle should be
$$(x-0)^2+(y-k)^2=5^2\Rightarrow x^2+y^2+k^2-2ky=25$$

and it passes through the point (3,2)
$$3^2+2^2+k^2-2k\cdot 2=25\Rightarrow k^2-4k-12=0\Rightarrow k=-2,6$$

But $$k=6$$ as the center is in the first quadrant

Putting this value in $$x^2+y^2+k^2-2ky=25$$, we get

$$x^2+y^2+36-12y=25\Rightarrow x^2+y^2-12y+11=0$$
Question 5
1 / -0

The axes are translated so that the new equation of the circle $$x^{2}+y^{2}-5x+2y-5=0$$ has no first degree terms. Then the new equation is

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

B
$$x^{2}+y^{2}=\dfrac {49}{4}$$

C
$$x^{2}+y^{2}=\dfrac {81}{16}$$

D
$$none\ of\ these$$

Solution

$$x^2+y^2-5x+2y-5=0$$
$$\sqrt {\dfrac {25}{4}+1+25}$$
$$=\sqrt {\dfrac {49}{4}}$$
$$ (5/2, -1)$$
$$x^2+y^2=\left (\sqrt {\dfrac {49}{4}}\right)^2$$
$$x^2+y^2=\dfrac {49}{4}$$
Question 6
1 / -0

A circle is concentric with circle $$x^{2}+ y^{2}-2x+4y-20=0$$. If perimeter of the semicircle is $$36$$ then the equation of the circle is :

A
$$x^{2}+y^{2}-2x-4y-44=0$$

B
$$(x-1) ^{2}+(y+2) ^{2}=(126/11) ^{2}$$

C
$$x^{2}+y^{2}-2x+4y-43=0$$

D
$$none\ of\ these$$

Solution

$$x^2+y^2-2x+4y-20=0$$
center $$(1, -2)$$
perimeter $$\Rightarrow 36=(\pi r+2r)$$
$$36=r(3.14+2)=$$
$$((x-1)+y+2)^2=\left(\dfrac{126}{11}\right)^2$$ or $$(7)^2$$ $$r=\dfrac{36}{5.14}=7$$
Question 7
1 / -0

The name of the conic represent by the equation $$x^2+y^2-2y+20x+10=0$$ is

A
a hyperbola

B
an ellipse

C
a parabola

D
circle

Solution

For a standard second degree equation $$ax^2+2hxy+by^2+2gx+2fy+c=0$$
to be a circle $$a=b\\h=0$$
Here $$a=b=1\\h=0$$
So The given equation is Circle
Question 8
1 / -0

The equation of the circle passing through the foci of the ellipes $${\frac{x}{{16}}^2} + {\frac{y}{{{9^{}}}}^2} = 1$$ and having centre at $$\left( {0,3} \right)$$ is

A
$${x^2} + {y^2} - 6y - 7 = 0$$

B
$${x^2} + {y^2} - 6y +7 = 0$$

C
$${x^2} + {y^2} - 6y - 5 = 0$$

D
$${x^2} + {y^2} - 6y + 5 = 0$$

Solution

R.E.F image
Equation of ellipse : $$ \frac{x^{2}}{16}+\frac{y^{2}}{9} = 1 $$
coordinate of foci are $$ (ac,o); (-ac,o) $$
$$ e = \sqrt{1-\frac{9}{16}} = \frac{\sqrt{7}}{4} $$
$$ a = 4 $$
co-ordinate of foci $$ = (\sqrt{7},0),(-\sqrt{7},0) $$
$$ R = \sqrt{7+9} = 4 $$
$$ (x-0)^{2}+(y-3)^{2} = 4^{2} $$
$$ x^{2}+y^{2}-6y+9 = 16 $$
$$ x^{2}+y^{2}-6y-7 = 0 $$
Question 9
1 / -0

A variable circle is drawn to touch the x-axis at the origin.The locus of the pole at the straight line $$6 x + m y + n = 0$$ w.r.t. the variable circle has the equation:-

A
$$x ( m y - n ) - e y ^ { 2 } = 0$$

B
$$x ( m y + n ) - e y ^ { 2 } = 0$$

C
$$x ( m y - n ) + \ell y ^ { 2 } = 0$$

D
none

Solution
Question 10
1 / -0

Equation of circles which touch both the axes and whose centres are at a distance of $$2\sqrt {2}$$ units from origin are

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

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

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

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

Solution

Since $$h=k$$ (radius )
$$\sqrt{h^{2}+k^{2}}=2\sqrt{2}$$
$$h^{2}+k^{2}=8$$
$$2h^{2}=8$$ (Taking circle $$1^{st}$$ Quad)
$$h=2$$
$$\therefore k=2$$ in
$$\therefore$$ center $$=(2,2)$$
Eqn. of circle having centre
$$(h,k)$$ & radius $$=r$$
$$(x-h)^{2}+(y-k)^{2}=r^{2}$$
$$\therefore (x-2)^{2}+ (y-2)^{2}=4$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Circles Test 18

Result

Circles Test 18

Score

-

Rank

-

SHARING IS CARING

Question 1

$${x}^{2}+{y}^{2}+2x-4y+6=0$$

$${x}^{2}+{y}^{2}-2x-4y+3=0$$

$${x}^{2}+{y}^{2}-2x+4y+8=0$$

$${x}^{2}+{y}^{2}-2x-4y+8=0$$

Solution

Question 2

$$(2,0),(-3,0)$$

$$(3,0),(4,0)$$

$$(1,0),(-1,0)$$

$$(1,0),(2,0)$$

Solution

Question 3

$$x^{2} + y^{2} - 4x - 8y - 97 = 0$$

$$x^{2} + y^{2} - 4x - 6y - 87 = 0$$

$$x^{2} + y^{2} - 2x - 8y - 95 = 0$$

None of these

Solution

Question 4

$${x^2} + {y^2} - 12y + 11 = 0$$

$${x^2} + {y^2} - 6y - 1 = 0$$

$${x^2} + {y^2} - 8y + 3 = 0$$

None of these

Solution

Question 5

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

$$x^{2}+y^{2}=\dfrac {49}{4}$$

$$x^{2}+y^{2}=\dfrac {81}{16}$$

$$none\ of\ these$$

Solution

Question 6

$$x^{2}+y^{2}-2x-4y-44=0$$

$$(x-1) ^{2}+(y+2) ^{2}=(126/11) ^{2}$$

$$x^{2}+y^{2}-2x+4y-43=0$$

$$none\ of\ these$$

Solution

Question 7

a hyperbola

an ellipse

a parabola

circle

Solution

Question 8

$${x^2} + {y^2} - 6y - 7 = 0$$

$${x^2} + {y^2} - 6y +7 = 0$$

$${x^2} + {y^2} - 6y - 5 = 0$$

$${x^2} + {y^2} - 6y + 5 = 0$$

Solution

Question 9

$$x ( m y - n ) - e y ^ { 2 } = 0$$

$$x ( m y + n ) - e y ^ { 2 } = 0$$

$$x ( m y - n ) + \ell y ^ { 2 } = 0$$

none

Solution

Question 10

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

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

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

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

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.