Circles Test 16

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Circles Test 16

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Weekly Quiz Competition

Question 1
1 / -0

The name of the conic represented by $$\sqrt{\dfrac{x}{a}}+\sqrt{\dfrac{y}{b}}=1$$ is

A
Circle

B
Parabola

C
Ellipse

D
Hyperbola

Solution

$$\sqrt{x/a}+\sqrt{y/b}=1$$

rearranging : $$\sqrt {y/b}=1-\sqrt{x/a}$$

squaring $$y/b=1+x/a-2\sqrt{x/a}$$

$$\Rightarrow 2\sqrt{x/a}=1+x/a-y/b$$

squaring again

$$4x/a=1+x^2/a^2+y^2/b^2+2x/a-\dfrac{2xy}{ab}-\dfrac{2y}{b}$$

Generally, when there is an equation of the form $$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$, if $$B^2-4AC=0$$, then the centre is one of the following:
(1) A parabola
(2) $$2$$ parallel lines ( certainly this is not the case)
(3) $$1$$ line ( certainly this is not the case)
(4) No curve ( again not the case).

$$\therefore \sqrt{x/a}+\sqrt{y/b}=1$$ is a part of a parabola
Question 2
1 / -0

Equation of circle having centre $$(5, 2)$$ and which passes through the point $$(1, -1)$$ is?

A
$$x^2+y^2-10x-4y-4=0$$

B
$$x^2+y^2+10x+4y+4=0$$

C
$$x^2+y^2-10x-4y-2=0$$

D
$$x^2+y^2-10x-4y+4=0$$

Solution
Question 3
1 / -0

Equation of the circle with radius 3 and centre as the point of intersection of the lines $$2x + 3y = 5, 2x - y = 1$$ is

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

B
$$x^2 + y^2 - 2x - 2y - 7 = 0 $$

C
$$x^2 + y^2 - 2x - 2y + 7 = 0$$

D
$$x^2 + y^2 + 9 = 0$$

Solution

radius $$= 3$$, $$2x+3y=5,2x-y=1$$

$$2x+3y=5$$
$$2x-y=1$$
___________
$$4y=4 \Rightarrow y=1$$ $$2x=2\Rightarrow x=1$$

Center of circle $$= (1,1)$$

$$\Rightarrow (x-1)^{2}+(y-1)^{2}=9$$

$$\Rightarrow x^{2}+1-2x+y^{2}+1-2y=9$$

$$\Rightarrow x^{2}+y^{2}-2x-2y-7=0$$
Question 4
1 / -0

A square is inscribed in the circle $$x^2 +y^2 - 4x - 6y-5 =0$$ whose sides are parallel to co-ordinate axes then vertices of square are

A
$$ (5, 0), (5, 6), (-1, 0), (-1, 6) $$

B
$$(5, 1), (5,-6), (-1, 1), (-1, 6)$$

C
$$ (5, 1), (5, 6), (-1, 0), (1, 6) $$

D
$$ (0, 5), (-6, 5), (0,-1),(6, 1) $$

Solution

$$x^2+y^2-4x-6y-5=0$$
$$(x-2)^2-4+(y-3)^2-9-5=0$$
$$(x-2)^2+(y-3)^2=18$$
centre $$(2,+3)$$ radius $$=\sqrt{18}=3\sqrt{2}$$
If a square is inscribe in circle,
$$\Rightarrow$$ diagonal of square =diameter
$$\sqrt{2}a=2(radius)$$
$$2=2\times 3\sqrt{2}$$
$$\Rightarrow a=6$$
$$\therefore$$ Side of the square is $$6$$
Since the side are parallel to $$x$$ and $$y$$
the vertices one equidistant from the centre.
$$(2+3,+3+3),(2-3,+3+3), (2-3,+3-3)$$ and $$(2+3,+3-3)$$
$$\Rightarrow (5,6), (-1,6), (-1,0),(5,0)$$
Question 5
1 / -0

If the lines $$3x-4y-7=0$$ and $$2x-3y-5=0$$ are two diameters of a circle of area 154 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

Let us find intersection of $$3x-4y-7=0$$
$$2x-3y-5=0$$
$$\dfrac {8y+14}{3}-3y-5=0$$
$$-9y+8y-15+19=0$$
$$y=-1$$
$$x=1$$
Area $$=l\ \pi \ r^2=154$$
$$\dfrac {22}{7}r^2 =154$$
$$r^2 =49$$
$$r=7$$ units
Circle: $$(x-1)^2 +(y+1)^2 =49$$
$$x^2 +y^2-2x+2y-47=0$$
$$C$$ is correct.
Question 6
1 / -0

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

A
$$(x+1)^2+(y+1)^2 = 49$$

B
$$(x-1)^2 + (y-1)^2=-49$$

C
$$(x-1)^2+(y+1)^2 = 49$$

D
$$(x+1)^2+(y-1)^2 = 49$$

Solution

Circle area $$=\pi r^{2}=154\Rightarrow r=7\ sq.units$$

intersection of diameter
$$(2x-3y=5)\times 3\\ (3x-4y=7)\times 2\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad \quad -y=1\Rightarrow y=-1$$

$$2x=5-3=2\Rightarrow x=1$$

eqn of circle $$\Rightarrow (x-1)^{2}+(y+1)^{2}=49$$
Question 7
1 / -0

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

A
Do not intersect

B
Are not orthogonal

C
Intersect orthogonally

D
Concentric

Solution

$$ S_1 = x^{2}+y^{2}+4y-4x+4 = 0 $$
$$ S_2 = x^2+y^2-4x-4y = 0 $$
$$ S_1 -S_2 = (x^2+y^2-4x+4y+4)-(x^2+y^2-4x-4y)$$
$$ = x^2+y^2-4x+4y-4-x^2-y^2+4x+4y$$
$$ = 8y-4 $$
$$ \Rightarrow 8y-4 = 0 \Rightarrow y = 1/2$$
from $$ S_1 = x^2+y^2+4y-4x+4 = 0 $$
$$ C_1$$ centre (2,-2)
$$ R_1 $$ Radius = $$ \sqrt{4+4-4} = 2 $$
similarly $$ S_2 = X^2+y^2-4x-4y = 0 $$
$$ C_2 $$ centre (2,2)
$$ R_2 $$ Radius $$= \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$$
Distance between centre of circles,
$$ \left | C_1 C_2 \right | = \sqrt{(2-2)^2+(2+2)^2 } = \sqrt{0+4^2} = \sqrt{4^2} = 4 $$
Now $$ R_1+R_2 = 2+2\sqrt{2} = 2 (1+\sqrt{2}) \neq \left | C_1 C_2\right |$$
$$ R_1 +R_2> $$
The condition for orthogonality $$ = 2gg'+2ff' = c+c'$$
$$ \Rightarrow 2(2)(2)+2(-2)(2) = 4+0 $$
$$ 8-8 = 0\neq 4 $$
So, no circles do not intersect orthogonality,
Option (b) is correct
Question 8
1 / -0

The set of values of p for which the power of a point $$(2,5)$$ is negative with respect to a circle $${ x }^{ 2 }+{ y }^{ 2 }-8x-12y+p=0$$ which neither touches the axes nor cuts them are

A
(36,57)

B
(36,47)

C
(37,47)

D
(16,47)

Solution

Power is negative $$\Rightarrow S_{1} < 0$$
$$\Rightarrow 2^{2}+5^{2}-16- 60+p < 0$$
$$\Rightarrow 4+25-76+p < 0$$
$$\Rightarrow p < 47$$
& square of x-coordinate of circle's
centre > square of radius (does not cut axes)
$$\Rightarrow g^{2} > g^{2}+f^{2}- c$$
$$\Rightarrow f^{2}- c < 0 \Rightarrow f^{2} < C$$
$$\Rightarrow 36 < p$$ similarly $$g^{2} < c$$
$$\Rightarrow 16 < p \Rightarrow p > 36$$
$$\Rightarrow p \in (36, 47)$$
Question 9
1 / -0

If the equation $$ax^{2}+2(a^{2}+ab-16)xy+by^{2}2ax+2by-\sqrt[4]{2}=0$$ represents a circle, the radius of the circle is

A
$$\sqrt{16 + \sqrt[4]{2}}$$.

B
$$\sqrt{24 + \sqrt[4]{2}}$$.

C
$$\sqrt{2}$$

D
$$\sqrt[4]{2}$$

Solution

$$a{x}^{2} + 2 \left( {a}^{2} + ab - 16 \right) xy + b {y}^{2} + 2ax + 2by - \sqrt[4]{2} = 0$$
If the above equation represents the equation of circle then-
$$a = b ..... \left( 1 \right)$$
$${a}^{2} + ab - 16 = 0$$
$$\Rightarrow {a}^{2} + {a}^{2} - 16 = 0$$
$$\Rightarrow 2{a}^{2} = 16$$
$$\Rightarrow {a}^{2} = 8$$
$$\therefore {b}^{2} = {a}^{2} = 8$$
Radius of circle $$\left( r \right) = \sqrt{{a}^{2} + {b}^{2} - c}$$
$$\Rightarrow r = \sqrt{8 + 8 + \sqrt[4]{2}}$$
$$\Rightarrow r = \sqrt{16 + \sqrt[4]{2}}$$
Hence the radius of the circle is $$\sqrt{16 + \sqrt[4]{2}}$$.
Question 10
1 / -0

A circle has radius $$3$$ units and its centre lies on the line $$y=x-1$$. if it passes through $$(7,3)$$, its equation

A
$$x^{2}+y^{2}-8x-14y=0$$

B
$$x^{2}+y^{2}-8x-6y-6=0$$

C
$$x^{2}+y^{2}-14x-12y-76=0$$

D
$$x^{2}+y^{2}+14x+12x-70=0$$

Solution

$${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={r}^{2}$$
Here $$\left(h,k\right)\rightarrow$$ centre
Equation of line through centre
$$k=h-1----(1)$$
& $$\left(x,y\right)$$ point on circle
So, given $$\left(x,y\right)= \left(7,3\right)$$
So, $${\left(7-h\right)}^{2}+{\left(3-h+1\right)}^{2}={3}^{2}$$
Using eqn $$(1)$$
We get,
$$28+{h}^{2}-11h=0$$
$$h=7$$ oR $$4$$
$$k=6$$ oR $$3$$
Equation of circle corresponding $$\left(7,6\right)$$
$${x}^{2}+{y}^{2}-14x-12y+76=0$$
Equation of circle corresponding $$\left(4,3\right)$$
$${x}^{2}+{y}^{2}-8x-6y+16=0$$

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Re-Attempt Weekly Quiz Competition

Circles Test 16

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Circles Test 16

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Question 1

Circle

Parabola

Ellipse

Hyperbola

Solution

Question 2

$$x^2+y^2-10x-4y-4=0$$

$$x^2+y^2+10x+4y+4=0$$

$$x^2+y^2-10x-4y-2=0$$

$$x^2+y^2-10x-4y+4=0$$

Solution

Question 3

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

$$x^2 + y^2 - 2x - 2y - 7 = 0 $$

$$x^2 + y^2 - 2x - 2y + 7 = 0$$

$$x^2 + y^2 + 9 = 0$$

Solution

Question 4

$$ (5, 0), (5, 6), (-1, 0), (-1, 6) $$

$$(5, 1), (5,-6), (-1, 1), (-1, 6)$$

$$ (5, 1), (5, 6), (-1, 0), (1, 6) $$

$$ (0, 5), (-6, 5), (0,-1),(6, 1) $$

Solution

Question 5

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

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

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

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

Solution

Question 6

$$(x+1)^2+(y+1)^2 = 49$$

$$(x-1)^2 + (y-1)^2=-49$$

$$(x-1)^2+(y+1)^2 = 49$$

$$(x+1)^2+(y-1)^2 = 49$$

Solution

Question 7

Do not intersect

Are not orthogonal

Intersect orthogonally

Concentric

Solution

Question 8

(36,57)

(36,47)

(37,47)

(16,47)

Solution

Question 9

$$\sqrt{16 + \sqrt[4]{2}}$$.

$$\sqrt{24 + \sqrt[4]{2}}$$.

$$\sqrt{2}$$

$$\sqrt[4]{2}$$

Solution

Question 10

$$x^{2}+y^{2}-8x-14y=0$$

$$x^{2}+y^{2}-8x-6y-6=0$$

$$x^{2}+y^{2}-14x-12y-76=0$$

$$x^{2}+y^{2}+14x+12x-70=0$$

Solution

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