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Circles Test 7

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Circles Test 7
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  • Question 1
    1 / -0
    If the centroid of an equilateral triangle is $$(1, 1)$$ and its one vertex is $$(-1, 2)$$ then the equation of its circumcircle is
    Solution
    Given centroid of an equilateral triangle is $$G(1,1)$$.
    We know that in an equilateral triangle, centroid, circumcenter and incenter are at the same point.
    So, the circumcenter is at $$G(1,1)$$.
    Given one vertex of equilateral triangle at $$A(-1,2)$$
    So, circumradius $$=AG=\sqrt { 5 } $$
    So, equation of circumcircle is
    $$(x-1)^2+(y-1)^2=5$$
    $$\Rightarrow x^{2}+y^{2}-2x-2y-3=0$$
  • Question 2
    1 / -0
    Two vertices of an equilateral triangle are $$(-1, 0)$$ and $$(1, 0)$$, and its third vertex lies above the $$x$$-axis. The equation of the circumcircle of the triangle is
    Solution
    Points $$A(-1,0)$$ and $$B(1,0)$$ lie on $$x$$-axis.

    Mid-point of $$AB$$ is $$(0,0)$$.

    Clearly, point $$C$$ lies on $$y$$-axis. Since, it lies above $$y$$-axis. 
    So, let the coordinates of point $$C$$ be $$(0, k)$$.

    Coordinates of centroid $$G$$ is $$\displaystyle \left (0,\dfrac{k}{3}\right)$$.
    We also know that circumcenter, centroid and incenter of the equilateral triangle coincides. 

    Hence, the coordinates of circumcenter is $$\displaystyle \left (0,\dfrac{k}{3}\right)$$.

    Since, $$ABC$$ is equilateral triangle,

    $$AB=AC$$

    $$\Rightarrow (AB)^2=(AC)^{2}$$

    $$\Rightarrow k=\sqrt{3}$$

    So, the coordinates of circumcenter is $$\displaystyle \left (0,\dfrac{1}{\sqrt{3}}\right)$$.

    Radius $$CG=\displaystyle \dfrac{2}{\sqrt{3}}$$

    Equation of circumcircle is

    $$(x-0)^{2}+\left (y-\dfrac{1}{\sqrt{3}}\right)^{2}=\displaystyle \dfrac{4}{3}$$

    $$\Rightarrow x^{2}+y^2-\dfrac{2}{\sqrt{3}}y-1=0$$

    $$\Rightarrow \displaystyle \sqrt{3}\left ( x^{2}+y^{2} \right )-2y-\sqrt{3}=0$$

  • Question 3
    1 / -0
    On the line joining the points $$A (0,4)$$ and $$B (3, 0)$$, a square $$ABCD $$ is constructed on the side of the line away from the origin. Equation of the circle having centre at $$C$$ and touching the axis of $$x$$ is
    Solution
    Let $$ \angle ABO = \theta $$, then $$\angle CBL= 90^{\circ}  -\theta, CL$$  being perpendicular to x-axis $$(Fig)$$. 
    The coordinates of $$C$$ are $$(OL, LC)$$ $$OL = OB + BL= 3 + 5 \sin \theta$$ $$= 3 + 5 \times  \left (\dfrac 45\right) = 7 $$ $$ CL = 5 \cos \theta = 5 \times \left (\dfrac 35\right) =3$$
    So, the coordinate of $$C$$ are $$(7, 3)$$ and the equation of the circle having $$C$$ as centre and touching $$x$$-axis is
    $$ (x -7)^{2} + (y-3)^{2} = (CL)^{2} = 9$$ 
    $$\Rightarrow  x^{2} + y^{2} -14x -6y + 49 = 0$$

  • Question 4
    1 / -0
    The equation of circle with origin as center and passing through the vertices of an equilateral triangle whose median is of length $$3a$$ is
    Solution
    The centroid of an equilateral triangle is the center of its circumcenter and the radius of the circle is the distance of any vertex from the centroid i.e. radius of the circle
    $$=$$ distance of centroid from any vertex
    $$\displaystyle =\frac { 2 }{ 3 } ($$Median$$)\displaystyle =\frac { 2 }{ 3 } (3a)=2a$$
    Hence, equation of circle whose center is $$(0,0)$$ and radius $$2a$$ is
    $${ \left( x-0 \right)  }^{ 2 }+{ \left( y-0 \right)  }^{ 2 }={ \left( 2a \right)  }^{ 2 }\Rightarrow { x }^{ 2 }+{ y }^{ 2 }=4{ a }^{ 2 }$$
  • Question 5
    1 / -0
    The lines $$2x-3y=5$$ and $$3x-4y=7$$ are diameters of a circle of area $$154\ sq.\ units$$. The equation of the circle is-
    Solution
    The centre of circle is the point of intersection of the given diameter $$2x-3y=5$$ and $$3x-4y=7$$
    which is $$\left( 1,-1 \right) $$ and if r is the radius of the circle ,then
    $$\displaystyle { \pi  }r^{ 2 }=154\Rightarrow { r }^{ 2 }=154\times \frac { 7 }{ 22 } \Rightarrow r=7$$
    Hence an equation of the required circle is
    $${ \left( x-1 \right)  }^{ 2 }+{ \left( y+1 \right)  }^{ 2 }={ 7 }^{ 2 }\\ \Rightarrow { x }^{ 2 }+{ y }^{ 2 }-2x+2y=47$$
  • Question 6
    1 / -0
    The lines $$2x -3y=5$$ and $$3x -4y =7$$ are the diameters of a circle of area $$154$$ square units. An equation of this circle is $$(\pi = 22/7)$$
    Solution
    The centre of the circle is the point of intersection of the given diameters
    $$2x- 3y = 5 $$ and $$3x- 4y=

    7,$$ which is $$(1,-1)$$
    and if $$r$$ is the radius of the circle,then

    $$\displaystyle \pi r^{2} = 154 \Rightarrow r^{2} = 154 \times

    \frac{7}{22} \Rightarrow  r = 7.$$
    Hence equation of the required

    circle is $$(x-1)^{2}+(y+ 1)^{2}=7^{2} \Rightarrow  x^{2}+y^{2}-

    2x+2y=47$$
  • Question 7
    1 / -0
    Find the equation of the circle whose centre is the point of intersection of the lines $$2x-3y+4=0$$ and $$3x+4y-5=0$$ and passes through the origin.
    Solution
    The point of intersection of the lines $$2x-3y+4=0$$ and $$3x+4y-5=0$$ is $$\left(\displaystyle-\frac{1}{17},\frac{22}{17}\right)$$
    Equation of circle whose centre is $$\left(\displaystyle-\frac{1}{17},\frac{22}{17}\right)$$ and passing through origin is
    $$x^2+y^2+\displaystyle\frac{2x}{17}-\displaystyle\frac{44y}{17}=0$$
    $$\Rightarrow 17(x^2+y^2)+2x-44y=0$$
    Hence, option A.
  • Question 8
    1 / -0
    The equation of the  circle passing through $$(3,6)$$ and whose centre is $$(2,-1)$$ is-
    Solution

    $$\textbf{Step -1: Find the radius of the circle.}$$

                    $$\text{As we know that, the standard equation of a circle is:}$$

                    $$(x-h)^2+(y-k)^2=r^2\text{ where }(h,k)\text{ is the centre and }r\text{ is the radius.}$$

                    $$\text{Here, the equation of the circle will be,}$$

                    $$(x-2)^2+(y+1)^2=r^2\ldots(i)$$

                    $$\text{and the point }(3,6)\text{ will satisfy this equation.}$$

                    $$\therefore (3-2)^2+(6+1)^2=r^2$$

                    $$\Rightarrow 1+49=r^2$$

                    $$\Rightarrow r^2=50$$

    $$\textbf{Step -2: Find the equation of the required circle.}$$

                    $$\text{On putting the value of }r^2\text{ in equation }(i),\text{ we get}$$

                    $$(x-2)^2+(y+1)^2=50$$

                    $$\Rightarrow x^2+4-4x+y^2+1+2y=50$$

                    $$\Rightarrow x^2+y^2-4x+2y=45$$

    $$\textbf{Hence, option A is correct.}$$

  • Question 9
    1 / -0
    The intercept on the line $$y=x$$ by the circle $${ x }^{ 2 }+{ y }^{ 2 }-2x=0$$ is $$AB$$. Equation of the circle with $$AB$$ as a diameter is
    Solution
    Equation of any circle passing through the point of intersection of 
    $${ x }^{ 2 }+{ y }^{ 2 }-2x=0$$ and $$y=x$$ is 
    $${ x }^{ 2 }+{ y }^{ 2 }-2x+\lambda \left( y-x \right) =0$$ $$\Rightarrow { x }^{ 2 }+{ y }^{ 2 }-\left( 2+\lambda  \right) x=\lambda y=0.$$
    Its centre is $$\displaystyle \left( \frac { 2+\lambda  }{ 2 } ,\frac { -\lambda  }{ 2 }  \right) .$$
    For $$AB$$ to be the diameter of the required circle, the centre must lie on $$AB,$$
     i.e., $$\displaystyle \frac { 2+\lambda  }{ 2 } =\frac { -\lambda  }{ 2 } \Rightarrow \lambda =-1.$$
    Thus, equation of required circle is $${ x }^{ 2 }+{ y }^{ 2 }-x-y=0.$$ 
  • Question 10
    1 / -0
    The normal at the point $$(3, 4)$$ on a circle cuts the circle again at the point $$(1, 2)$$. Then the equation of the circle is -
    Solution
    We know the normal of a circle passes through it's center.
    So the given points $$(3,4)$$ and $$(1,2)$$ are end point of the diameter.
    and  the centre of the circle is mid point of the given points, $$\displaystyle \equiv \left( 2,3 \right) $$
    and Radius $$=\displaystyle \frac { \sqrt { 4+4 }  }{ 2 } =\frac { 2\sqrt{ 2 }  }{ 2 } =\sqrt { 2 } $$
    Hence required circle is, $$\displaystyle (x-2{ ) }^{ 2 }+(y-3{ ) }^{ 2 }=2$$
    $$\Rightarrow \displaystyle { x }^{ 2 }+{ y }^{ 2 }-4x-6y+11=0$$
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