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Circles Test 23

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Circles Test 23
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  • Question 1
    1 / -0
    The equation of the circle, the end points of whose diameter are the centre of the circles $$x^{2}+y^{2}+6x-14y=1$$ and $$x^{2}+y^{2}-4x+10y=2$$ is 
    Solution
    The general equation of circle is $$x^2 + y^2 + 2gx + 2fy + c = 0$$
    where,
    Center = $$(-g, -f)$$
    Radius = $$\sqrt{g^2 + f^2 - c}$$

    So,
    For $$x^2 + y^2 + 6x - 14y =1$$,
    Center = $$(-3,7)$$

    And For $$x^2 + y^2 - 4x + 10y = 2$$,
    Center = $$(2,-5)$$

    Since centers of given circles are diametric ends of new circle,
    For new circle,
    Center = $$(\dfrac{-3+2}{2}, \dfrac{7-5}{2}) = (\dfrac{-1}{2}, 1)$$
    Radius = $$\dfrac{\sqrt{(-3-2)^2 + (7+5)^2}}{2} = \dfrac{13}{2}$$

    Hence, Equation of new circle
    $$\rightarrow (x+\dfrac{1}{2})^2 + (y-1)^2 = \dfrac{169}{4}$$
    $$\rightarrow x^2 + y^2 + x -2y - 41 = 0 $$
  • Question 2
    1 / -0
    The equation of a circle with centre at $$(2,-3)$$ and the circumference is $$10 \pi$$ units is 
    Solution
    For a circle with center (a,b) and radius r,
    Equation is $$(x-a)^2 + (y-b)^2 = r^2$$

    First let's find radius of circle,
    $$\rightarrow $$ Circumference =  $$2\times \pi \times r$$
    $$\rightarrow $$ $$10 \times \pi$$ = $$2 \times \pi \times r$$
    $$\rightarrow $$ $$r = 5$$

    Equation of circle,
    $$\rightarrow$$ $$ (x-2)^2 + (y+3)^2 = 5^2$$
    $$\rightarrow $$ $$ x^2 + y^2 -4x + 6y - 12 = 0$$
  • Question 3
    1 / -0
    Equation having circle centre $$(5, 2)$$ and which passes through the point $$(1, -1)$$ is 
    Solution
    equation of circle,

    $$(x-h)^2+(y-k)^2=r^2$$

    $$C=(5,2),P=(1,-1)$$

    $$\Rightarrow (1-5)^2+(-1-2)^2=r^2$$

    $$16+9=r^2$$

    $$\therefore r=5$$

    Now,

    $$(x-5)^2+(y-2)^2=5^2$$

    $$x^2+25-10x+y^2-4-4y=25$$

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

  • Question 4
    1 / -0
    The parametric equation of the circle $$x^{2}+y^{2}+x+\sqrt {3}y=0$$ are
    Solution
    The parametric equation of circle with radius r and center (a,b) is,
    x = a + rcos$$\theta$$, y = b + rsin$$\theta$$

    The general equation on the other side is,
    $$(x-a)^2 + (y-b)^2 = r^2$$

    Let's find the radius using equation $$x^2 + y^2 + x + \sqrt{3}y = 0$$
    We can rewrite equation as, $$(x+\dfrac{1}{2})^2 + (y+\dfrac{\sqrt{3}}{2})^2 = 1$$  

    Comparing this with general equation, we get,
    $$a = \dfrac{-1}{2}, b= \dfrac{-\sqrt{3}}{2}, r = 1$$

    Hence, the parametric equation of circle is,
    x = $$\dfrac{-1}{2} + cos\theta$$, y = $$\dfrac{-\sqrt{3}}{2} + sin\theta$$
  • Question 5
    1 / -0
    If two vertices of an equilateral triangle are $$A (-a, 0)$$ and $$B(a, 0), a > 0$$ and the third vertex C lies above x-axis then the equation of the circumcircle of $$\Delta ABC$$ is
    Solution

  • Question 6
    1 / -0
    The equation of the circle of radius $$5$$ with centre on  x-axis and passing through the point $$(2,3)$$ is
    Solution
    For a circle with center (a,b) and radius r,
    Equation is $$(x-a)^2 + (y-b)^2 = r^2$$

    Let the center of given circle be (k,0).
    Then, equation of circle,
    $$\rightarrow$$ $$(x-k)^2 + (y-0)^2 = 5^2$$

    Since circle passes through (2,3),
    $$\rightarrow$$ $$(2-k)^2 + 3^2 = 5^2$$
    Solving we get, k = {6,-2}

    So, Possible equations are,
    (i) $$(x-6)^2 + y^2 = 25$$ $$\rightarrow$$ $$x^2 + y^2 -12x +11 = 0$$
    (ii) $$(x+2)^2 + y^2 = 25$$ $$\rightarrow$$ $$x^2 + y^2 + 4x -21 = 0$$ 

    Hence, (A) and (B) are correct answers.
  • Question 7
    1 / -0
    The equation(s) of the circle(s) which pass through the ends of the common chords of two circles $$2x^{2}+2y^{2}+8x+4y-7=0$$ and $$x^{2}+y^{2}-8x-4y-5=0$$ and touch the line $$x=7$$ is (are) :
  • Question 8
    1 / -0
    The equation of the circle which touches the axes of $$y$$ at the origin and passing through $$(3,4)$$ is
    Solution

  • Question 9
    1 / -0
    The equation of the circle with centre at $$(4, 3)$$ and touching the line $$5x-12y-10=0$$ is?
    Solution

  • Question 10
    1 / -0
    Equation of circle touching the line $$x + y = 4$$ at $$( 1, 3)$$ and intersecting the circle $${ x }^{ 2 }+{ y }^{ 2 }=4$$ orthogonally is
    Solution
    The general equation of circle touching the line lx + my + n = 0 at given point $$(x_1,y_1)$$ is $$(x-x_1)^2+(y-y_1)^2 + \lambda(lx+my+n)=0$$ where $$\lambda \in R$$

    For two circles with equation, 
    $$x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$$ and $$x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$$,
    Condition of orthogonality $$\rightarrow 2g_1g_2 + 2f_1f_2 = c_1 + c_2$$

    So, for given condition equation will be,
    $$\rightarrow (x-1)^2 + (y-3)^2 + \lambda (x+y-4) = 0$$
    $$\rightarrow x^2 + y^2 +(\lambda -2)x + (\lambda-6)y +( 10-4\lambda) = 0$$

    Using condition of orthoganility,
    $$2(2-\lambda)(0) + 2(6-\lambda)(0) = -4 + 10 - 4\lambda$$
    $$\rightarrow \lambda = \dfrac{3}{2}$$

    So, Equation of circle,
    $$\rightarrow x^2 + y^2 - \dfrac{1}{2}x -\dfrac{9}{2}y + 4 = 0$$
    $$\rightarrow 2x^2 + 2y^2 -x-9y + 8 =0$$

    Hence, (D) is the correct answer.
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