Circles Test 21

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Circles Test 21

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Weekly Quiz Competition

Question 1
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The equation of the circle of a radius $$5$$ in the first quadrant which touches the x-axis and the line $$3x-4y=0$$ is

A
$$x^{2}+y^{2}-24x-y-25=0$$

B
$$x^{2}+y^{2}-20x-12y+144=0$$

C
$$x^{2}+y^{2}-16x-18y+64=0$$

D
$$x^{2}+y^{2}-30x-10y+255=0$$

Solution

Hence, Option (D) is the correct answer.
Question 2
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Equation of circle touching $$x=0, y=0$$ and $$x=4$$ is

A
$$4(x^{2}+y^{2})-16x-16y+16=0$$

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

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

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

Solution

$$\textbf{Step -1: Finding the centre and radius from the given values.}$$
$$\text{We are given that}$$ $$(h,k)=(2,2),r=2$$
$$\textbf{Step -2: Finding the equation of the circle}$$
$$\text{Equation of a circle whose centre is}$$ $$(h,k)$$ $$\text{and radius is}$$ $$r$$ $$\text{is}$$
$$\Rightarrow (x-h)^2+(y-k)^2=r^2$$
$$\Rightarrow (x-2)^2+(y-2)^2=2^2$$
$$\Rightarrow x^2-4x+4+y^2+4y+4=4$$
$$\Rightarrow x^2+y^2-4x-4y=-4$$
$$\text{Multiplying throughout by}$$ $$4,$$
$$\Rightarrow 4(x^2+y^2)-16x-16y+16=0$$
$$\textbf{ Hence, correct option is A.}$$
Question 3
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The circle $$x ^ { 2 } + y ^ { 2 } = 4 x + 8 y + 5$$ intersects the line $$3 x - 4 y = m$$ at two distinct points if :-

A
$$- 85 < m < - 35$$

B
$$- 35 < \mathrm { m } < 15$$

C
$$15 < \mathrm { m } < 65$$

D
$$35 < m < 85$$

Solution

circle $$x^{2}+y^{2}-4x-8y-5=0$$

standard circle : $$ ax^{2}+2nxy+by^{2}+2gx+2fy+c=0$$
on comparing -

center $$(2,4)$$

Radius$$=\sqrt{4+16+5}=\sqrt{25}=5$$

If the circle is interacting line $$3x-4y=m$$ at distinct
points:

$$\Rightarrow $$ Length of perpendicular from center to
line $$ <$$ radius

$$\Rightarrow \left | \dfrac{6-16-m}{5} \right |<25$$

$$\Rightarrow \left | 10+m \right |< 25$$

$$\Rightarrow -25< m+10< 25$$

$$\Rightarrow -35< m< 15$$
Question 4
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Find the equation of a circle whose center is (2,-1) and radius is 3

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

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

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

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

Solution

Equation of circle with center $$(h,k)$$ and radius $$r$$ is written as
$$(x-h)^2+(y-k)^2=r^2$$

then,
Equation of circle with center $$(2,-1)$$ and radius $$3$$ is written as
$$\Rightarrow (x-2)^2+(y+1)^2=3^2$$
$$\Rightarrow x^2+4-4x+y^2+1+2y=9$$
$$\therefore x^2+y^2-4x+2y-4=0$$
Question 5
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The equation of the circle passing through $$(4, 1)$$ and $$(6, 5)$$ and having center of the line

A
$$x^2+y^2-6x-8y+15=0$$

B
$$x^2+y^2-6x-8y-15=0$$

C
$$x^2+y^2-6x-8y-11=0$$

D
$$x^2+y^2-6x-8y+11=0$$

Solution

Let the equation of the required circle be $$(x-h)^2+(y-k)^2=r^2$$ …………$$(1)$$
It is given that the circle passes through $$(4, 1)$$ and $$(6, 5)$$
$$\therefore (4-h)^2+(1-k)^2=r^2$$ and $$(6-h)^2+(5-k)^2=r^2$$
$$\Rightarrow (4-h)^2+(1-k)^2=(6-h)^2+(5-k)^2$$
on expanding we get,
$$16-8h+h^2+1-2k+k^2=36-12h+h^2+25-10k+k^2$$
$$16-8h+1-2k=36-12h+25-10k$$
$$4h+8k=44$$.
Dividing throughout by $$4$$, we get,
$$h+2k=1$$ ……….$$(2)$$
It is given that the centre of the circle lies on the line $$4x+y=16$$
$$\therefore 4h+k=16$$ ………….$$(3)$$
Solving $$(2)$$ & $$(3)$$
$$4h+k=16$$
$$4h+8k=44$$
_____________
$$-7k=-25$$
$$k=4$$
_____________
Substitute for k in equation $$(1)$$
$$h+2(4)=11$$
$$h=3$$
$$\therefore h=3$$ & $$k=4$$ putting in equation $$(1)$$
$$(4-3)^2+(1-4)^2=r^2$$
$$r=\sqrt{10}$$
Hence, $$(x-3)^2+(y-4)^2=(\sqrt{10})^2$$
$$\therefore x^2+y^2-6x-8y+15=0$$.
Question 6
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The equation of the tangent to the curve y = 2sinx + sin2x at $$x=\frac { \pi }{ 3 } $$ on it is

A
$$(2,3)$$

B
$$y+\sqrt { 3 } =0$$

C
$$2t-3=0$$

D
$$2y-3\sqrt { 3 } =0$$

Solution

$${ (x-2) }^{ 2 }+{ (y-3) }^{ 2 }=0\\ \Rightarrow { x }^{ 2 }+4-4x+{ y }^{ 2 }+9-6y=0\\ { x }^{ 2 }+{ y }^{ 2 }-4x+6y+13=0$$
radius$$=\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } $$
$$g=-2 \quad f=-3 \quad c=13$$
$$r=\sqrt { { (-2) }^{ 2 }+{ (-3) }^{ 2 }-13 } =\sqrt { 13-13 } =0$$
radius of the circle is zero .
Hence diameter$$=0$$
i.e., it is a point circle with centre at $$(2,3)$$.
Question 7
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$${ \cos }^{ 4 }\dfrac { \pi }{ 8 } +{ \cos }^{ 4 }\dfrac { 3\pi }{ 8 } +{ \cos }^{ 4 }\dfrac { 5\pi }{ 8 } +{ \cos }^{ 4 }\dfrac { 7\pi }{ 8 } =$$

A
$$\cfrac { 1 }{ 2 } $$

B
$$\cfrac { 3 }{ 2 } $$

C
$$\cfrac { 1 }{ 4 } $$

D
$$\cfrac { 3 }{ 4 } $$

Solution

$$\textbf{Step 1: Simplify the given expression.}$$

$$\text{Consider, } \cos^4\dfrac{\pi}{8} + \cos^4\dfrac{3\pi}{8} + \cos^4\dfrac{5\pi}{8} + \cos^4\dfrac{7\pi}{8}$$

$$= \cos^4\dfrac{\pi}{8} + \cos^4\left(\dfrac{\pi}{2}-\dfrac{\pi}{8}\right) + \cos^4 \left ( \dfrac{\pi}{2} + \dfrac{\pi}{8} \right ) + \cos^4 \left ( \pi - \dfrac{\pi}{8} \right )$$

$$= \cos^4\dfrac{\pi}{8} + \sin^4\dfrac{\pi}{8} + \left (-\sin\dfrac{\pi}{8} \right )^4 + \left ( -cos\dfrac{\pi}{8} \right )^4$$

$$= 2\left [ \cos^4\dfrac{\pi}{8} + \sin^4\dfrac{\pi}{8} \right]$$

$$\textbf{Step 2: Use multiple angle formula to get final result. }$$

$$= 2 \left \{ \left ( \cos^2\dfrac{\pi}{8} + \sin^2\dfrac{\pi}{8} \right )^2 - 2\cos^2\dfrac{\pi}{8} \cdot \sin^2\dfrac{\pi}{8} \right \}$$

$$= 2 \left ( 1^2 - \dfrac{1}{2} \left ( 2\sin\dfrac{\pi}{8}\cos\dfrac{\pi}{8} \right )^2 \right )$$

$$= 2\left [ 1^2 - \dfrac{1}{2} \times \left ( \sin\dfrac{\pi}{4} \right )^2 \right ]$$ $$[\because \boldsymbol{\sin 2A=2\sin A\cos A}]$$

$$= 2\left ( 1^2 - \dfrac{1}{2} \times \left ( \dfrac{1}{\sqrt 2} \right )^2 \right )$$

$$= 2\left [1^2 - \dfrac{1}{4} \right ]=\dfrac{3}{2}$$

$$\textbf{Hence, Option 'B' is correct.}$$
Question 8
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The line $$y=mx+c$$ cut the circle $${x}^{2}+{y}^{2}={a}^{2}$$ in the distinct point $$A$$ and $$B$$. Equation of the circle having minimum radius that an be drawn through the points $$A$$ and $$B$$ is

A
$$\left( 1+{ m }^{ 2 } \right) \left( { x }^{ 2 }+{ y }^{ 2 }-{ a }^{ 2 } \right) +2c\left( y-mx-c \right) =0$$

B
$$\left( 1+{ m }^{ 2 } \right) \left( { x }^{ 2 }+{ y }^{ 2 }-{ a }^{ 2 } \right) +c\left( y-mx-c \right) =0$$

C
$$\left( 1+{ m }^{ 2 } \right) \left( { x }^{ 2 }+{ y }^{ 2 }-{ a }^{ 2 } \right) -2c\left( y-mx-c \right) =0$$

D
$$\left( 1+{ m }^{ 2 } \right) \left( { x }^{ 2 }+{ y }^{ 2 }-{ 2a }^{ 2 } \right) -2c\left( y-mx-c \right) =0$$

Solution
Question 9
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Three sides of a triangle have the equations $$L_{r} = y - m_r x - C_{r} = 0; r = 1, 2, 3$$. Then $$\lambda L_{2}L_{3} + \mu L_{3}L_{1} + \gamma L_{1}L_{2} = 0$$. where $$\lambda \neq 0, \mu \neq 0, \gamma \neq 0$$, is the equation of circumcircle of triangle if

A
$$\lambda (m_{2} + m_{3}) + \mu (m_{3} + m_{1}) + \gamma (m_{1} + m_{2}) = 0$$

B
$$\lambda (m_{2}m_{3} - 1) + \mu (m_{3}m_{1} - 1) + \gamma (m_{1}m_{2} - 1) = 0$$

C
Both $$(a)$$ and $$(b)$$ hold together

D
None of these

Solution
Question 10
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The equation $$14x^{2}-4xy+11y^{2}-44x-58y+71=0$$ represents

A
a circle

B
an ellipse

C
a hyperbola

D
a rectangular hyperbola

Solution