Circles Test 26

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Circles Test 26

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Question 1
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A thin rod of length $$l$$ in the shape of a semicircle is pivoted at one of its ends such that it is free to oscillate in its own plane. The frequency $$f$$ of small oscillations of the semicircular rod is :

A
$$\dfrac{1}{2\pi}\sqrt{\dfrac{g\pi}{2l}}$$

B
$$\dfrac{1}{2\pi}\sqrt{\dfrac{g\sqrt{\pi^2+4}}{2l}}$$

C
$$\dfrac{1}{2\pi}\sqrt{\dfrac{g\sqrt{\pi+2}}{l}}$$

D
$$\dfrac{1}{2\pi}\sqrt{\dfrac{g\sqrt{\pi^2+1}}{2 \pi l}}$$

Solution

We know that $$T=2\pi \sqrt{\dfrac{I}{mgl_{eff}}}$$
$$I=2mR^2$$
So,
$$l_{eff}=\sqrt{R^2+(\dfrac{2R}{\pi})^2}=\dfrac{R}{\pi}\sqrt{\pi^2+4}$$
On subsituting these value in T, we get
$$T=2\pi \sqrt{\dfrac{2mR^2}{mg\dfrac{R}{\pi}\sqrt{\pi^2+4}}}=2\pi \sqrt{\dfrac{2\pi R}{g\sqrt{\pi^2+4}}}$$
We know that $$R=\dfrac{l}{\pi}$$
Therefore,
$$T=2\pi \sqrt{\dfrac{2\pi l}{\pi g\sqrt{\pi^2+4}}}=2\pi \sqrt{\dfrac{2l}{g\sqrt{\pi^2+4}}}$$
Frequency $$f=\dfrac{1}{T}=\dfrac{1}{2\pi}\sqrt{\dfrac{g\sqrt{\pi^2+4}}{2l}}$$
Question 2
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Directions For Questions

Circle touching a line $$L=0$$ at a point $$\left({x}_{1},{y}_{1}\right)$$ on it is
$${\left(x-{x}_{1}\right)}^{2}+{\left(y-{y}_{1}\right)}^{2}+\lambda L=0,\lambda\in R$$
Circle through the two points $$A\left({x}_{1},{y}_{1}\right)$$ and $$B\left({x}_{2},{y}_{2}\right)$$ is
$$\left(x-{x}_{1}\right)\left(x-{x}_{2}\right)+\left(y-{y}_{1}\right)\left(y-{y}_{2}\right)+\lambda L=0 , \lambda \in R$$
where $$L=0$$ is the equation of the line $$AB$$
On the basis of the above information,answer the following questions:

...view full instructions

From the point $$A\left(0,3\right)$$ on the circle $${x}^{2}-4x+{\left(y-3\right)}^{2}=0$$ a chord $$AB$$ is drawn and extended to a point $$M$$ such that $$AM=2AB$$.The locus is

A
$${x}^{2}+{y}^{2}-8x-6y+9=0$$

B
$${x}^{2}+{y}^{2}+8x-6y+9=0$$

C
$${x}^{2}+{y}^{2}-8x+6y+9=0$$

D
$${x}^{2}+{y}^{2}+8x+6y+9=0$$

Solution

$${x}^{2}-4x+{\left(y-3\right)}^{2}=0$$
Add $$4$$ both sides of the above equation, we get
$$\Rightarrow {x}^{2}-4x+4+{\left(y-3\right)}^{2}=4$$
$$\Rightarrow {\left(x-2\right)}^{2}+{\left(y-3\right)}^{2}=4$$
Take $$B=\left(2+2\cos\theta,3+2\sin\theta\right)$$
$$\therefore 2+2\cos\theta=\dfrac{x+0}{2} , 3+2\sin\theta=\dfrac{y+3}{2}$$
or $$2\cos\theta=\dfrac{x}{2}-2, 2\sin\theta=\dfrac{y+3}{2}-3$$
or $$2\cos\theta=\dfrac{x-4}{2}, 2\sin\theta=\dfrac{y-3}{2}$$
Eliminating $$\theta$$ we get
$${\left(2\cos\theta\right)}^{2}+{\left(2\sin\theta\right)}^{2}={\left(\dfrac{x-4}{2}\right)}^{2}+{\left(\dfrac{y-3}{2}\right)}^{2}$$
$$4={\left(\dfrac{x-4}{2}\right)}^{2}+{\left(\dfrac{y-3}{2}\right)}^{2}$$
$${\left(x-4\right)}^{2}+{\left(y-3\right)}^{2}=16$$
On simplifying , we get
$${x}^{2}+{y}^{2}-8x-6y+9=0$$
Question 3
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If $$(\alpha, \beta)$$ is a point on the circle whose centre is on the x-axis and which touches the line x + y = 0 at (2, -2), then the greatest value of $$\alpha$$ is

A
$$ 4 - \sqrt{2}$$

B
6

C
$$ 4 + \sqrt{2}$$

D
None of these

Solution

Given: $$(\alpha,\beta)$$ is a point on the circle whose centre is an the x-axis and which touches the line $$x+y=0$$ AT $$(2,-2)$$

Let $$(a,o)$$ is the centre C and P is $$(2,-2)$$

which is given

then $$\angle COP =45^0$$

Since the equation of $$OP$$ is $$x+y=0$$

thus also then $$\angle OCP=45^0$$

$$\therefore OP=2\sqrt?{2}=CP$$

Here, $$OC=4$$

The point on the circle with the greatest coordinate is $$B$$.

$$\alpha=OB$$

$$\alpha=OC+CB$$

$$\alpha=4+2\sqrt{2}$$

Hence, greatest value of $$\alpha=4+2\sqrt{2}$$

$$\therefore$$ Option C is correct.
Question 4
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Circles are drawn passing through the origin $$O$$ to intersect the coordinate axes at point $$P$$ and $$Q$$ such that $$m$$. $$PO+n.OQ=k$$, then the fixed point satisfy all of them, is given by

A
$$(m,n)$$

B
$$\dfrac{m^{2}}{k},\dfrac{n^{2}}{k}$$

C
$$\dfrac{mk}{m^{2}+n^{2}},\dfrac{nk}{m^{2}+n^{2}}$$

D
$$(k,k)$$

Solution

$${x}^{2}+{y}^{2}+2gx+2fy+c=0$$
$$C=0$$
$${x}^{2}+{y}^{2}+2gx+2fy=0$$
$${x}^{2}+2g\left(x\right)=0$$
$$x\left(x+2g\right)=0$$
$${x}^{2}+{y}^{2}+2gx+2fy=0\quad \quad H.O.P+nOQ=k$$
For$$\left(m, n\right)\quad {m}^{2}+{n}^{2}+2gm+2fn\quad \quad H\left(-2g\right)+n\left(-2f\right)=k$$
$${m}^{2}+{n}^{2}-k\neq0\quad \quad 2mg-2n7=-k$$
for$$\left(\dfrac{mk}{{m}^{2}+{n}^{2}}- \dfrac{nk}{{m}^{2}+{n}^{2}}\right)=$$
$$\left( \dfrac { { m }^{ 2 }{ k }^{ 2 } }{ { \left( { m }^{ 2 }{ +k }^{ 2 } \right) }^{ 2 } } \quad \dfrac { n^{ 2 }{ k }^{ 2 } }{ { \left( { m }^{ 2 }{ +n }^{ 2 } \right) }^{ 2 } } \right) +2g\left( \dfrac { mk }{ { m }^{ 2 }{ +n }^{ 2 } } \right) +2f\left( \dfrac { nk }{ { m }^{ 2 }{ +n }^{ 2 } } \right) \\ $$
$$=\dfrac { { k }^{ 2 }\left( { m }^{ 2 }+{ n }^{ 2 } \right) +k\left( { m }^{ 2 }+{ n }^{ 2 } \right) \left( -k \right) }{ { \left( { m }^{ 2 }+{ n }^{ 2 } \right) }^{ 2 } } $$
$$\Rightarrow \left(\dfrac{mk}{{m}^{2}+{n}^{2}}, \dfrac{nk}{{m}^{2}+{n}^{2}}\right)$$
Question 5
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The centre of a circle is $$(2, -3)$$ and the circumference is $$10\pi$$. Then, the equation of the circle is

A
$${x}^{2}+{y}^{2}+4x+6y+12=0$$

B
$${x}^{2}+{y}^{2}-4x+6y+12=0$$

C
$${x}^{2}+{y}^{2}-4x+6y-12=0$$

D
$${x}^{2}+{y}^{2}-4x-6y-12=0$$

Solution

The circumference of the circle is given as $$10\pi $$.

$$C = 2\pi r$$

$$10\pi = 2\pi r$$

$$r = \frac{{10\pi }}{{2\pi }}$$

$$r = 5$$

The general form of the equation is,

$${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$$

The center of the circle is given as $$\left( {h,k} \right) = \left( {2, - 3} \right)$$

$${\left( {x - 2} \right)^2} + {\left( {y - \left( { - 3} \right)} \right)^2} = {\left( 5 \right)^2}$$

$${\left( {x - 2} \right)^2} + {\left( {y + 3} \right)^2} = 25$$

$${x^2} + 4 - 4x + {y^2} + 9 + 6y = 25$$

$${x^2} + {y^2} - 4x + 6y + 13 = 25$$

$${x^2} + {y^2} - 4x + 6y - 25 + 13 = 0$$

$${x^2} + {y^2} - 4x + 6y - 12 = 0$$

Therefore, the equation of the circle is,

$${x^2} + {y^2} - 4x + 6y - 12 = 0$$
Question 6
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A conic $$C$$ passes through the points $$(2,4)$$ and is such that the segment of any of its tangents at any point contained between the co-ordinate axis is biscected at the point of tangency. Let $$S$$ denotes circle described on the foci $${F_1}$$ and $${F_2}$$ of the conic $$C$$ as diameter.
Equation of the circle $$S$$ is

A
$${x^2} + {y^2} = 16$$

B
$${x^2} + {y^2} = 8$$

C
$${x^2} + {y^2} = 32$$

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

Solution
Question 7
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The centre of a circle passing through the point $$(0,0),(1,0)$$ and touching the circle $$x^{2}+y^{2}=9$$ is ?

A
$$\left(\dfrac {3}{2},\dfrac {1}{2}\right)$$

B
$$\left(\dfrac {1}{2},\dfrac {3}{2}\right)$$

C
$$\left(\dfrac {1}{2},\dfrac {1}{2}\right)$$

D
None of these

Solution

Let the circle be$$x^{2}+y^{2}+2gx+2fy+c=0$$
$$\because$$ It passes through $$(0,0)$$
$$\Rightarrow 0+c=0$$
$$\Rightarrow c=0$$
It also passes through $$(1,0)$$
$$\Rightarrow 1+0+2g+0+c=0$$
$$\Rightarrow g=\dfrac{-c-1}{2}=\dfrac{-1}{2}$$
$$\therefore$$Circle $$\rightarrow x^{2}+y^{2}-x+2fy=0$$
Centre$$\rightarrow (\dfrac{1}{2},-f)$$
Other circle : $$x^{2}+y^{2}=9$$
Centre \rightarrow (0,0)$$
Radius $$=3$$
$$\because Circles touches internally.
$$\therefore $$ Difference between radius = Distance between centres.
Radius of smaller circle $$=\sqrt{g^{2}+f^{2}-0}=\sqrt{\dfrac{1}{4}+f^{2}}$$
$$\Rightarrow 3-\sqrt{\dfrac{1}{4}+f^{2}}=\sqrt{(-g-0)^{2}+(-f-0)^{2}}$$
$$\Rightarrow 3-\sqrt{\dfrac{1 }{4}+f^{2}}=\sqrt{\dfrac{1}{4}+f^{2}}$$
$$\Rightarrow 2\sqrt{\dfrac{1}{4}+f^{2}}=3$$
$$\Rightarrow \dfrac{1}{4}+f^{2}=\dfrac{9}{4}$$
$$\Rightarrow f^{2}=\dfrac{8}{4}=2$$
$$\Rightarrow f=\pm \sqrt{2}$$
Hence centre $$\rightarrow (-g,-f)$$
$$\Rightarrow(\dfrac{1}{2},\pm\sqrt{2})$$
$$\therefore$$None of the options is correct.
Question 8
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If the centroid of an equilateral triangle $$(1,1)$$ and its one vertex is $$(-1,2)$$ , then the equation of the circumcircle is

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

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

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

D
$${(x + 2)^2} + {y^2} = 5$$

Solution
Question 9
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The equation of the circle having the lines $$x^2 + 2xy + 3x + 6y = 0$$ as its normals and having size just sufficient to contain the circle $$x (x - 4) + y(y - 3) = 0$$ is

A
$$x^2 + y^2 + 3x - 6y - 40 = 0$$

B
$$x^2 + y^2 + 6x - 3y - 45 = 0$$

C
$$x^2 + y^2 + 8x + 4y - 20 = 0$$

D
$$x^2 + y^2 + 4x + 8y + 20 =0$$

Solution

Pair of lines:
$$x^{2}+2xy+3x+6y=0$$
$$\Rightarrow x(x+2y)+3(x+2y)=0$$
$$\Rightarrow (x+2y)(x+3)=0$$
$$\Rightarrow x+2y=0$$
and $$x+3=0$$
Are the two normals and their point of intersection must be the centre of the circle.
$$\Rightarrow x=-3$$
and $$y=\dfrac{-x}{2}=\dfrac{3}{2}$$
$$\Rightarrow (-3,\dfrac{3}{2}) $$ is the centre of required circle.
($$\because x(x-4)+y(y-3)=0$$ is the diametric form of the circle. Hence, $$(0,0)$$ and $$(4,3)$$ are the diametric end points.)
$$\Rightarrow$$ Centre $$\rightarrow (\dfrac{0+4}{2},\dfrac{0+3}{2})\rightarrow (2,\dfrac{3}{2})$$
radius $$=\dfrac{1}{2}(\sqrt{16+9})$$
$$=\dfrac{5}{2}$$
The required circle with centre $$(-3,\dfrac{3}{2})$$ is just sufficient to contain the circle
$$x(x-4)+y(y-3)=0$$
$$\therefore$$ radius of required circle
=distance between $$(-3,\dfrac{3}{2})$$ and $$(2,\dfrac{3}{2})$$ $$+$$ radius of given circle.
$$=\sqrt{(-3-2)^{2}+0}+\dfrac{5}{2}$$
$$=5+\dfrac{5}{2}=\dfrac{15}{2}$$
$$\therefore$$ Equation of required circle:
$$\Rightarrow (x+3)^{2}+(y-\dfrac{3}{2})^{2}=\dfrac{225}{4}$$
$$\Rightarrow x^{2}+y^{2}+9+\dfrac{9}{4}+6x-3y=\dfrac{225}{4}$$
$$\Rightarrow x^{2}+y^{2}+6x-3y+9+\dfrac{9}{4}-\dfrac{225}{4}=0$$
$$\Rightarrow x^{2}+y^{2}+6x-3y+9-\dfrac{216}{4}=0$$
$$\Rightarrow x^{2}+y^{2}+6x-3y+9-54=0$$
$$\Rightarrow x^{2}+y^{2}+6x-3y-45=0$$
$$\therefore B)$$ Answer.
Question 10
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If the equation $$\frac { \lambda ( x + 1 ) ^ { 2 } } { 3 } + \frac { ( y + 2 ) ^ { 2 } } { 4 } = 1$$ represents a circle then $$\lambda =$$

A
1

B
$$\frac { 3 } { 4 }$$

C
0

D
$$- \frac { 3 } { 4 }$$

Solution