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NCERT Exemplar · Class 11 Mathematics Conic Sections

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EXERCISE 11.3 1–10 (part 1 of 6)

  1. Exercise 1

    Find the equation of the circle which touches the both axes in first quadrant and whose radius is a\displaystyle a.

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    NCERT’s answer
    \(\displaystyle x^2+y^2-2 a x-2 a y+a^2=0\)
    The centre is at distance \(\displaystyle a\) from each axis, so it is \(\displaystyle (a, a)\).NCERT_Solution_Class11_Maths_Exemplar_Ch11_Ex11-3_Q1\[(x-a)^2 + (y-a)^2 = a^2 \] \[x^2 + y^2 - 2ax - 2ay + a^2 = 0 \]Answer: \(\displaystyle x^2 + y^2 - 2ax - 2ay + a^2 = 0\)

  2. Exercise 2

    Show that the point (x,y)\displaystyle (x, y) given by x=2at1+t2\displaystyle x=\frac{2 a t}{1+t^2} and y=a(1−t2)1+t2\displaystyle y=\frac{a\left(1-t^2\right)}{1+t^2} lies on a circle for all real values of t\displaystyle t such that −1≤t≤1\displaystyle -1 \leq \mathrm{t} \leq 1 where a\displaystyle a is any given real numbers.

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    \[x^2 + y^2 = \frac{4a^2t^2 + a^2(1-t^2)^2}{(1+t^2)^2} \] \[= \frac{a^2\left(4t^2 + 1 - 2t^2 + t^4\right)}{(1+t^2)^2} \] \[= \frac{a^2(1+t^2)^2}{(1+t^2)^2} = a^2 \]NCERT_Solution_Class11_Maths_Exemplar_Ch11_Ex11-3_Q2Answer: every such point satisfies \(\displaystyle x^2 + y^2 = a^2\), the circle with centre \(\displaystyle (0, 0)\) and radius \(\displaystyle |a|\).

  3. Exercise 3

    If a circle passes through the point (0,0)(a,0),(0,b)\displaystyle (0,0)(a, 0),(0, b) then find the coordinates of its centre.

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    NCERT’s answer
    \(\displaystyle \left(\frac{a}{2}, \frac{b}{2}\right)\)
    Take the circle \(\displaystyle x^2 + y^2 + 2gx + 2fy + c = 0\), with \(\displaystyle a, b \neq 0\).\[c = 0 \quad \text{(through } (0,0)\text{)} \] \[a^2 + 2ga = 0 \Rightarrow g = -\tfrac{a}{2} \quad \text{(through } (a,0)\text{)} \] \[b^2 + 2fb = 0 \Rightarrow f = -\tfrac{b}{2} \quad \text{(through } (0,b)\text{)} \] \[\text{centre} = (-g, -f) = \left(\tfrac{a}{2}, \tfrac{b}{2}\right) \]NCERT_Solution_Class11_Maths_Exemplar_Ch11_Ex11-3_Q3Answer: \(\displaystyle \left(\dfrac{a}{2}, \dfrac{b}{2}\right)\)

  4. Exercise 4

    Find the equation of the circle which touches x\displaystyle x-axis and whose centre is (1,2)\displaystyle (1,2).

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    NCERT’s answer
    \(\displaystyle x^2+y^2-2 x-4 y+1=0\)
    The circle touches the \(\displaystyle x\)-axis, so its radius is the distance of \(\displaystyle (1, 2)\) from that axis.\[r = 2 \]NCERT_Solution_Class11_Maths_Exemplar_Ch11_Ex11-3_Q4\[(x-1)^2 + (y-2)^2 = 2^2 \] \[x^2 + y^2 - 2x - 4y + 1 = 0 \]Answer: \(\displaystyle x^2 + y^2 - 2x - 4y + 1 = 0\)

  5. Exercise 5

    If the lines 3x−4y+4=0\displaystyle 3 x-4 y+4=0 and 6x−8y−7=0\displaystyle 6 x-8 y-7=0 are tangents to a circle, then find the radius of the circle. [Hint: Distance between given parallel lines gives the diameter of the circle.]

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    NCERT’s answer
    \(\displaystyle \frac{3}{4}\)
    The lines are parallel; write the second as \(\displaystyle 3x - 4y - \tfrac{7}{2} = 0\).\[d = \frac{\left|4 - \left(-\tfrac{7}{2}\right)\right|}{\sqrt{3^2 + 4^2}} = \frac{15/2}{5} = \frac{3}{2} \]The circle lies between the two tangents, so \(\displaystyle d\) is its diameter.NCERT_Solution_Class11_Maths_Exemplar_Ch11_Ex11-3_Q5\[r = \frac{d}{2} = \frac{3}{4} \]Answer: \(\displaystyle r = \dfrac{3}{4}\)

  6. Exercise 6

    Find the equation of a circle which touches both the axes and the line 3x−4y+8=0\displaystyle 3 x-4 y+8=0 and lies in the third quadrant. [Hint: Let a\displaystyle a be the radius of the circle, then (−a,−a)\displaystyle (-a,-a) will be centre and perpendicular distance from the centre to the given line gives the radius of the circle.]

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    NCERT’s answer
    \(\displaystyle x^2+y^2+4 x+4 y+4=0\)
    The circle lies in the third quadrant and touches both axes, so the centre is \(\displaystyle (-a, -a)\) with radius \(\displaystyle a > 0\). Its distance from the line equals \(\displaystyle a\).\[\frac{|3(-a) - 4(-a) + 8|}{\sqrt{3^2 + 4^2}} = a \] \[\frac{|a + 8|}{5} = a \] \[a + 8 = 5a \quad (a + 8 > 0) \] \[a = 2 \]NCERT_Solution_Class11_Maths_Exemplar_Ch11_Ex11-3_Q6\[(x+2)^2 + (y+2)^2 = 2^2 \] \[x^2 + y^2 + 4x + 4y + 4 = 0 \]Answer: \(\displaystyle x^2 + y^2 + 4x + 4y + 4 = 0\)

  7. Exercise 7

    If one end of a diameter of the circle x2+y2−4x−6y+11=0\displaystyle x^2+y^2-4 x-6 y+11=0 is (3,4)\displaystyle (3, 4), then find the coordinate of the other end of the diameter.

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    NCERT’s answer
    \(\displaystyle (1,2)\)
    \[\text{centre} = \left(-\tfrac{-4}{2},\, -\tfrac{-6}{2}\right) = (2, 3) \]The centre is the midpoint of a diameter. Let the other end be \(\displaystyle (x, y)\).\[\frac{3 + x}{2} = 2, \qquad \frac{4 + y}{2} = 3 \] \[x = 1, \qquad y = 2 \]NCERT_Solution_Class11_Maths_Exemplar_Ch11_Ex11-3_Q7Answer: \(\displaystyle (1, 2)\)

  8. Exercise 8

    Find the equation of the circle having (1,−2)\displaystyle (1,-2) as its centre and passing through 3x+y=14,2x+5y=18\displaystyle 3 x+y=14,2 x+5 y=18

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    NCERT’s answer
    \(\displaystyle x^2+y^2-2 x+4 y-20=0\)
    The circle passes through the point where the two lines meet. Solve them.\[3x + y = 14 \Rightarrow y = 14 - 3x \] \[2x + 5(14 - 3x) = 18 \Rightarrow -13x = -52 \Rightarrow x = 4 \] \[y = 14 - 3(4) = 2 \]The point is \(\displaystyle P(4, 2)\); the centre is \(\displaystyle (1, -2)\).\[r^2 = (4-1)^2 + (2+2)^2 = 25 \]NCERT_Solution_Class11_Maths_Exemplar_Ch11_Ex11-3_Q8\[(x-1)^2 + (y+2)^2 = 25 \] \[x^2 + y^2 - 2x + 4y - 20 = 0 \]Answer: \(\displaystyle x^2 + y^2 - 2x + 4y - 20 = 0\)

  9. Exercise 9

    If the line y=3x+k\displaystyle y=\sqrt{3} x+k touches the circle x2+y2=16\displaystyle x^2+y^2=16, then find the value of k\displaystyle k. [Hint: Equate perpendicular distance from the centre of the circle to its radius].

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    NCERT’s answer
    \(\displaystyle k \pm 8\)
    Tangent means the distance from the centre \(\displaystyle (0,0)\) to the line equals the radius \(\displaystyle 4\). \[\sqrt{3}\,x - y + k = 0 \] \[\frac{|k|}{\sqrt{(\sqrt{3})^2 + (-1)^2}} = 4 \] \[\frac{|k|}{2} = 4 \] \[|k| = 8 \] Answer: \(\displaystyle k = \pm 8\)

  10. Exercise 10

    Find the equation of a circle concentric with the circle x2+y2−6x+12y+15=0\displaystyle x^2+y^2-6 x+12 y+15=0 and has double of its area. [Hint: concentric circles have the same centre.]

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    NCERT’s answer
    \(\displaystyle x^2+y^2-6 x+12 y-15=0\)
    \[x^2 + y^2 - 6x + 12y + 15 = 0 \] \[(x-3)^2 + (y+6)^2 = 9 + 36 - 15 = 30 \] Centre \(\displaystyle (3,-6)\), \(\displaystyle r^2 = 30\). Concentric, so the centre is the same. \[\pi R^2 = 2\pi r^2 \] \[R^2 = 60 \] \[(x-3)^2 + (y+6)^2 = 60 \] \[x^2 + y^2 - 6x + 12y + 45 - 60 = 0 \] Answer: \(\displaystyle x^2 + y^2 - 6x + 12y - 15 = 0\)