Self Studies
Selfstudy
Selfstudy

Circles Test 14

Result Self Studies

Circles Test 14
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

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.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0

    Directions For Questions

    Equation of a circle is  $$S={x}^{2}+{y}^{2}+2gx+2fy+c$$
    Its notation is $${S}_{1}={x}_{1}x+{y}_{1}y+g\left(x+{x}_{1}\right)+f\left({y}_{1}+y\right)+c$$
    $${S}_{11}={x}_{1}^{2}+{y}_{1}^{2}+2g{x}_{1}+2f{y}_{1}+c$$
    $$\left(i\right)$$Location of $$P\left({x}_{1},{y}_{1}\right):$$
    $$P$$ lies inside the circle $$S=0$$, if $${S}_{11}<0$$
    $$P$$ lies outside the circle $$S=0$$ if $${S}_{11}>0$$
    $$P$$ lies on the circle $$S=0,$$ if $${S}_{11}=0$$
    $$\left(ii\right)$$Tangent at $$P\left({x}_{1},{y}_{1}\right)$$ on the circle $$S=0,$$ is $${S}_{1}=0$$
    $$\left(iii\right)$$ Length of the tangent from the point $$\left({x}_{1},{y}_{1}\right)$$ to the circle $$S=0$$ is $$\sqrt{{S}_{11}}$$
    $$\left(iv\right)$$Pair of tangents $$PQ,PR$$ from $$P\left({x}_{1},{y}_{1}\right)$$ is $${S}_{1}^{2}={S}_{11}S$$
    $$\left(v\right)$$Chord of contact $$QR$$ of tangents from $$P\left({x}_{1},{y}_{1}\right)$$ is $${S}_{1}=0$$
    $$\left(vi\right)$$Chord of $$S=0$$ with midpoint $$\left({x}_{1},{y}_{1}\right)$$ is $${S}_{1}={S}_{11}$$
    Based on the above information,answer the following questions:

    ...view full instructions

    The equation of the circle whose radius is $$5$$units and which touches the circle, $${x}^{2}+{y}^{2}-2x-4y-20=0$$ at the point $$\left(5,5\right)$$ is
    Solution
    The given circle is $${x}^{2}+{y}^{2}-2x-4y-20=0$$ has its centre 
    $$-2g=-2$$ or $$g=1$$
    $$-2f=-4$$ or $$f=2$$$
    Thus, the centre is at $$C\left(1,2\right)$$
    and radius$$=r=\sqrt{{g}^{2}+{f}^{2}-c}=\sqrt{{1}^{2}+{2}^{2}-\left(-20\right)}=5$$units
    We have $${O}_{1}=\left(1,2\right)$$ and Let $${O}_{2}=\left(\alpha,\beta\right)$$ be the centres $$A\left(5,5\right)$$ is the mid-point of $${O}_{1}{O}_{2}$$
    $$\therefore \dfrac{1+\alpha}{2}=5, \dfrac{2+\beta}{2}=5$$
    On simplifying ,we get
    $$\alpha=9,\beta=8$$
    The circle is $${\left(x-9\right)}^{2}+{\left(y-8\right)}^{2}={5}^{2}$$
    or $${x}^{2}+{y}^{2}-18x-16y+120=0$$
  • Question 2
    1 / -0

    In the $$xy$$ plane,  the segment with end points$$(3,8)$$ and $$(
    5,2)$$ is the diameter of the circle. The  point $$(k,10)$$ lies on the circle for:

    Solution
    The centre of the given circle is,

    $$C\left( x,y \right)=\left( \dfrac{3+5}{2},\dfrac{8+2}{2} \right)=\left( 4,5 \right)$$

     

    The length of the diameter of the circle is,

    $$D=\sqrt{{{\left( 5-3 \right)}^{2}}+{{\left( 2-8 \right)}^{2}}}=2\sqrt{10}\ units$$

     

    Therefore,

    Radius $$=\sqrt{10}\ units$$

     

    Therefore, the equation of the required circle is,

    $${{\left( x-4 \right)}^{2}}+{{\left( y-5 \right)}^{2}}=10$$

     

    Since, the point $$\left( k,10 \right)$$ lies on the circle, we have

     $$ {{\left( k-4 \right)}^{2}}+{{\left( 10-5 \right)}^{2}}=10 $$

     $$ {{k}^{2}}-8k+16+25=10 $$

     $$ {{k}^{2}}-8k+31=0 $$

     

    This equation has no real roots. Therefore, no such value of $$k$$ exists.
  • Question 3
    1 / -0
    Find the equation of the circle which passes through the points $$(2,-2)$$ and $$(3,4)$$. And whose centre lies on the line $$x+y=2$$.
    Solution
    As $$ (2,-2)$$ passes through the circle.
    $$ { x }^{ 2 }+{ y }^{ 2 }+2gx+2fy+c=0\\ 4+4+4g-4f+c=0\\ 4g+4f+8+c=0\longrightarrow (1)$$
     As $$ (3,4)$$ passes through the circle
    $$ 25+6g+8f+c=0\longrightarrow (2)$$
     As center of a circle $$ (-g,-f)$$
    $$ -g-f=2\\ g+f=-2\longrightarrow (3)\\ 4g-4f+8+c=0\\ 4\left( g-f+2+\cfrac { c }{ 4 }  \right) =0\\ g-f+2+\cfrac { c }{ 4 } =0\\ -4=\cfrac { c }{ 4 } \\ c=-16\\ 4g-4f-8=0\times 2\\ 8g-8f-16=0\\ 6g+8f+9=0\times 1\\ 6g+8f+9=0\\ 2g=-7\\ g=\cfrac { -7 }{ 2 } \\ f=\cfrac { 3 }{ 2 } $$
    Equation of circle
    $$ { x }^{ 2 }+{ y }^{ 2 }+2\times \cfrac { 7 }{ 2 } x-2\times \cfrac { 3 }{ 2 } y-16=0$$ 
    $${ x }^{ 2 }+{ y }^{ 2 }+7x-3y-16=0$$
  • Question 4
    1 / -0

    Directions For Questions

    If $$7{l}^{2}-9{m}^{2}+8l+1=0$$ and we have to find the equation of circle having $$lx+my+1=0$$ is a tangent and we can adjust given condition as $$16{l}^{2}+8l+1=9\left({l}^{2}+{m}^{2}\right)$$
    or $${\left(4l+1\right)}^{2}=9\left({l}^{2}+{m}^{2}\right)\Rightarrow \dfrac{\left|4l+1\right|}{\sqrt{\left({l}^{2}+{m}^{2}\right)}}=3$$
    Center of circle$$=\left(4,0\right)$$ and radius$$=3$$ when any two non-parallel lines touching a circle, then centre of circle lies on angle bisector of lines.

    ...view full instructions

    If $$16{m}^{2}-8l-1=0,$$ then equation of the circle having $$lx+my+1=0$$ is a tangent is
    Solution
    $$\because 16{m}^{2}=8l+1$$
    $$\Rightarrow 16\left({l}^{2}+{m}^{2}\right)=16{l}^{2}+8l+1={\left(4l+1\right)}^{2}$$
    or $$4\sqrt{\left({l}^{2}+{m}^{2}\right)}=\left|4l+1\right|$$
    or $$\dfrac{\left|4l+1\right|}{\sqrt{\left({l}^{2}+{m}^{2}\right)}}=4$$
    $$\therefore$$ Centre$$\equiv\left(4,0\right)$$ and radius$$\equiv4$$
    Equation of circle is $${\left(x-4\right)}^{2}+{\left(y-0\right)}^{2}={4}^{2}$$
    $$\Rightarrow {x}^{2}+16-8x+{y}^{2}=16$$
    or $$ {x}^{2}-8x+{y}^{2}=0$$ is the required equation of the circle
  • Question 5
    1 / -0
    A circle of radius $$2$$ lies in the first quadrant and touches both the axes of co-ordinates. Then the equation of the circle with centre $$(6, 5)$$ and touching the above circle externally is
    Solution
    If $$(h,k)$$ is the center and the radius is $$r$$ then the equation of the circle is given by $$(x-h)^2+(y-k)^2=r^2$$

    Given that The center of the circle $$(h,k)=(6,5)$$ and the radius $$r=2$$

    Therefore, the equation of the circle is  $$(x-6)^2+(y-5)^2=4$$
  • Question 6
    1 / -0
    The equation of the circle having normal at $$(3, 3)$$ as $$y = x$$ and passing through $$(2, 2)$$ is:
    Solution
    $${ x }^{ 2 }+y^{ 2 }+2gx+2fy+c=0\\ 2x+2y\cfrac { dy }{ dx } +2g+2f\cfrac { dy }{ dx } =0.\\ \cfrac { dy }{ dx } =\cfrac { -y-x }{ y+f. } \\ \cfrac { dy }{ dx } \quad at\quad (3,3).\\ \Rightarrow \cfrac { dy }{ dx } =\cfrac { -y-3 }{ 3+f } \\ Slope\quad of\quad normal\quad at\quad this\quad point\quad =1\\ \Rightarrow -\cfrac { (g+3) }{ f+3 } (1)\quad =-1\\ \Rightarrow g+3=f+3.\\ \Rightarrow g=f.\\ \Rightarrow { x }^{ 2 }+y^{ 2 }+2gx+2fy+c=0\\ Circle\quad passes\quad through\quad (3,3)\& (2,2).\\ \Rightarrow g+g+bg+bg+c=0\\ \Rightarrow 12g=-18-c.\\ \Rightarrow 4+4+4g+4g+c=0.\\ 8g=-8-c.\\ -12g=-18-c\\ -4g=10\\ g=\cfrac { -5 }{ 2 } =f.\\ C=-8(g+1)\\ =-8(\cfrac { -5 }{ 2 } +1)\\ =-8\times  \cfrac { -3 }{ 2 } \\ =+12\\ \therefore \quad Equation\quad :\\ { x }^{ 2 }+y^{ 2 }-5x-5y+12=0.\\ Ans.$$
  • Question 7
    1 / -0
    If the equation $$\dfrac{\lambda (x+1)^2}{3}+\dfrac{(y+2)^2}{4}=1$$ represents a circle then $$\lambda = ?$$
    Solution
    $$\dfrac{\lambda (x+1)^{2}}{3}+\dfrac{(y+2)^{2}}{4}=1$$

    for a circle of $$a(x-\alpha )^{2}+6(y-\beta )^{2}=1$$ then $$(a=6)$$
    hence
    $$\dfrac{\lambda }{3}=\dfrac{1}{4}$$
    $$\lambda =\dfrac{3}{4}$$
  • Question 8
    1 / -0
    Coordinates of centre and radius of the circle $$(x-3)^2+(y+4)^2=25$$ are respectively
    Solution
    $$(x-3)^{2}+(y+4)^{2}=25$$
    $$(x-3)^{2}+(y-(-4))^{2}-(5)^{2}$$
    $$(x-4)^{2}+(y-k)^{2}-(r)^{2}$$
    $$r=5 \,\,(h,k)= (3,-4)$$
  • Question 9
    1 / -0
    If the equation $$\dfrac { \lambda { \left( x+1 \right)  }^{ 2 } }{ 3 } +\dfrac { { \left( y+2 \right)  }^{ 2 } }{ 4 }=1$$ represents a circle then $$\lambda=$$
    Solution
    in a circle$$x^2,y^2$$ coefficients must be same [it is the condition for the equation to be circle]

    therefore

    $$\dfrac{\lambda}{3}=\dfrac{1}{4}$$

    $$\Rightarrow \lambda=\dfrac 34$$ 

    option B 
  • Question 10
    1 / -0
    The equation of the circle touches y axis and having radius $$2$$ units and centre is $$(-2, -3)$$?
    Solution

Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now