SolveItNCERT · CBSE Boards

NCERT Solutions · Class 11 Mathematics Conic Sections

70 exercises · 70 still being checked

EXERCISE 10.1 1–10 (part 1 of 8)

  1. In each of the following Exercises $\displaystyle 1$ to $\displaystyle 5$, find the equation of the circle with

    Exercise 1

    centre \(\displaystyle (0,2)\) and radius $\displaystyle 2$

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    $\displaystyle x^{2}+y^{2}-4 y=0$
    Standard form of a circle. The circle with centre \(\displaystyle (h,k)\) and radius \(\displaystyle r\) has equation \[(x-h)^{2}+(y-k)^{2}=r^{2}.\]Here \(\displaystyle h=0,\ k=2,\ r=2\), so \[(x-0)^{2}+(y-2)^{2}=2^{2}\] \[x^{2}+y^{2}-4y+4=4\]\(\displaystyle x^{2}+y^{2}-4y=0\)

  2. Exercise 2

    centre $\displaystyle (-2,3)$ and radius $\displaystyle 4$

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    $\displaystyle x^{2}+y^{2}+4 x-6 y-3=0$
    Standard form of a circle. The circle with centre \(\displaystyle (h,k)\) and radius \(\displaystyle r\) has equation \[(x-h)^{2}+(y-k)^{2}=r^{2}.\]Here \(\displaystyle h=-2,\ k=3,\ r=4\), so \[(x+2)^{2}+(y-3)^{2}=4^{2}\] \[x^{2}+4x+4+y^{2}-6y+9=16\]\(\displaystyle x^{2}+y^{2}+4x-6y-3=0\)

  3. Exercise 3

    centre \(\displaystyle \left(\frac{1}{2}, \frac{1}{4}\right)\) and radius \(\displaystyle \frac{1}{12}\)

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    $\displaystyle 36 x^{2}+36 y^{2}-36 x-18 y+11=0$
    Standard form of a circle. The circle with centre \(\displaystyle (h,k)\) and radius \(\displaystyle r\) has equation \[(x-h)^{2}+(y-k)^{2}=r^{2}.\]Here \(\displaystyle h=\tfrac12,\ k=\tfrac14,\ r=\tfrac1{12}\), so \[\left(x-\frac12\right)^{2}+\left(y-\frac14\right)^{2}=\frac{1}{144}.\]Expanding, \[x^{2}-x+\frac14+y^{2}-\frac{y}{2}+\frac1{16}=\frac1{144}.\]Multiply throughout by \(\displaystyle 144\): \[144x^{2}-144x+36+144y^{2}-72y+9=1\] \[144x^{2}+144y^{2}-144x-72y+44=0,\] and dividing by \(\displaystyle 4\):\(\displaystyle 36x^{2}+36y^{2}-36x-18y+11=0\), i.e. \(\displaystyle \left(x-\tfrac12\right)^{2}+\left(y-\tfrac14\right)^{2}=\tfrac1{144}\)

  4. Exercise 4

    centre \(\displaystyle (1,1)\) and radius \(\displaystyle \sqrt{2}\)

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    $\displaystyle x^{2}+y^{2}-2 x-2 y=0$
    Standard form of a circle. The circle with centre \(\displaystyle (h,k)\) and radius \(\displaystyle r\) has equation \[(x-h)^{2}+(y-k)^{2}=r^{2}.\]Here \(\displaystyle h=1,\ k=1,\ r=\sqrt2\), so \[(x-1)^{2}+(y-1)^{2}=\left(\sqrt2\right)^{2}=2\] \[x^{2}-2x+1+y^{2}-2y+1=2\]\(\displaystyle x^{2}+y^{2}-2x-2y=0\)

  5. Exercise 5

    centre \(\displaystyle (-a,-b)\) and radius \(\displaystyle \sqrt{a^{2}-b^{2}}\).

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    $\displaystyle x^{2}+y^{2}+2 a x+2 b y+2 b^{2}=0$
    Standard form of a circle. The circle with centre \(\displaystyle (h,k)\) and radius \(\displaystyle r\) has equation \[(x-h)^{2}+(y-k)^{2}=r^{2}.\]Here \(\displaystyle h=-a,\ k=-b,\ r=\sqrt{a^{2}-b^{2}}\), so \[(x+a)^{2}+(y+b)^{2}=a^{2}-b^{2}\] \[x^{2}+2ax+a^{2}+y^{2}+2by+b^{2}=a^{2}-b^{2}\]\(\displaystyle x^{2}+y^{2}+2ax+2by+2b^{2}=0\)(This needs \(\displaystyle a^{2}\ge b^{2}\) for the radius to be real.)
  6. In each of the following Exercises $\displaystyle 6$ to $\displaystyle 9$, find the centre and radius of the circles.

    Exercise 6

    \(\displaystyle (x+5)^{2}+(y-3)^{2}=36\)

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    $\displaystyle c(-5,3), r=6$
    Reading off the standard form. In \(\displaystyle (x-h)^{2}+(y-k)^{2}=r^{2}\) the centre is \(\displaystyle (h,k)\) and the radius is \(\displaystyle r\).Write the given equation in that shape: \[(x+5)^{2}+(y-3)^{2}=36\ \Longrightarrow\ \bigl(x-(-5)\bigr)^{2}+(y-3)^{2}=6^{2}.\]So \(\displaystyle h=-5,\ k=3,\ r=6\).Centre \(\displaystyle (-5,3)\), radius \(\displaystyle 6\).

  7. Exercise 7

    \(\displaystyle x^{2}+y^{2}-4 x-8 y-45=0\)

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    $\displaystyle c(2,4), r=\sqrt{65}$
    Completing the square. Group the \(\displaystyle x\)-terms and the \(\displaystyle y\)-terms and complete each square; the equation then reads \(\displaystyle (x-h)^{2}+(y-k)^{2}=r^{2}\).\[x^{2}-4x+y^{2}-8y=45\] \[(x^{2}-4x+4)+(y^{2}-8y+16)=45+4+16\] \[(x-2)^{2}+(y-4)^{2}=65\]So \(\displaystyle h=2,\ k=4,\ r=\sqrt{65}\).Centre \(\displaystyle (2,4)\), radius \(\displaystyle \sqrt{65}\).

  8. Exercise 8

    \(\displaystyle x^{2}+y^{2}-8 x+10 y-12=0\)

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    $\displaystyle c(4,-5), r=\sqrt{53}$
    Completing the square.\[x^{2}-8x+y^{2}+10y=12\] \[(x^{2}-8x+16)+(y^{2}+10y+25)=12+16+25\] \[(x-4)^{2}+(y+5)^{2}=53\]So \(\displaystyle h=4,\ k=-5,\ r=\sqrt{53}\).Centre \(\displaystyle (4,-5)\), radius \(\displaystyle \sqrt{53}\).

  9. Exercise 9

    \(\displaystyle 2 x^{2}+2 y^{2}-x=0\)

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    $\displaystyle c\left(\frac{1}{4}, 0\right) ; r=\frac{1}{4}$
    Completing the square. First make the coefficients of \(\displaystyle x^{2}\) and \(\displaystyle y^{2}\) equal to \(\displaystyle 1\) by dividing by \(\displaystyle 2\): \[x^{2}+y^{2}-\frac{x}{2}=0.\]Now complete the square in \(\displaystyle x\) (there is no \(\displaystyle y\)-term): \[\left(x^{2}-\frac{x}{2}+\frac1{16}\right)+y^{2}=\frac1{16}\] \[\left(x-\frac14\right)^{2}+(y-0)^{2}=\left(\frac14\right)^{2}\]Centre \(\displaystyle \left(\dfrac14,\,0\right)\), radius \(\displaystyle \dfrac14\).

  10. Exercise 10

    Find the equation of the circle passing through the points \(\displaystyle (4,1)\) and \(\displaystyle (6,5)\) and whose centre is on the line \(\displaystyle 4 x+y=16\).

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    $\displaystyle x^{2}+y^{2}-6 x-8 y+15=0$
    Locating the centre from equal radii. Let the centre be \(\displaystyle C(h,k)\). Since the centre lies on \(\displaystyle 4x+y=16\), \[4h+k=16. \tag{1}\]\(\displaystyle C\) is equidistant from the two points on the circle, so \(\displaystyle CA^{2}=CB^{2}\) with \(\displaystyle A(4,1)\), \(\displaystyle B(6,5)\): \[(h-4)^{2}+(k-1)^{2}=(h-6)^{2}+(k-5)^{2}\] \[h^{2}-8h+16+k^{2}-2k+1=h^{2}-12h+36+k^{2}-10k+25\] \[4h+8k=44\ \Longrightarrow\ h+2k=11. \tag{2}\]From ($\displaystyle 1$), \(\displaystyle k=16-4h\); substituting in ($\displaystyle 2$): \(\displaystyle h+32-8h=11\Rightarrow -7h=-21\Rightarrow h=3\), hence \(\displaystyle k=4\).Radius: \(\displaystyle r^{2}=(3-4)^{2}+(4-1)^{2}=1+9=10\) (and \(\displaystyle (3-6)^{2}+(4-5)^{2}=9+1=10\), which checks).\[(x-3)^{2}+(y-4)^{2}=10\]\(\displaystyle x^{2}+y^{2}-6x-8y+15=0\)