Chapter 10: Conic Sections

Ans: The equation of a circle with centre (h,k) and radius (r) is given as

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

It is given that centre $(h,k)=(0,2)$ and radius $(r)=2.$

Therefore, the equation of the circle is

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

$\begin{align}   & {{x}^{2}}+{{y}^{2}}+4-4y=4 \\  & {{x}^{2}}+{{y}^{2}}-4y=0 \\ \end{align}$

Ans: The equation of a circle with centre (h,k) and radius (r) is given as

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

It is given that centre (h,k)=(-2,3) and radius (r)=4.

Therefore, the equation of the circle is

${{(x+2)}^{2}}+{{(y-3)}^{2}}={{(4)}^{2}}$

$\begin{align}  & {{x}^{2}}+4x+4+{{y}^{2}}-6y+9=16 \\  & {{x}^{2}}+{{y}^{2}}+4x-6y-3=0 \\ \end{align}$

Ans: The equation of a circle with centre (h,k) and radius (r) is given as

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

It is given that centre $(h,k)=\left( \dfrac{1}{2},\dfrac{1}{4} \right)$ and radius $(r)=\left( \dfrac{1}{12} \right).$

Therefore, the equation of the circle is

${{\left( x-\dfrac{1}{2} \right)}^{2}}+{{\left( y-\dfrac{1}{4} \right)}^{2}}={{\left( \dfrac{1}{12} \right)}^{2}}$

${{x}^{2}}-x+\dfrac{1}{4}+{{y}^{2}}-\dfrac{y}{2}+\dfrac{1}{16}=\dfrac{1}{144}$

${{x}^{2}}-x+\dfrac{1}{4}+{{y}^{2}}-\dfrac{y}{2}+\dfrac{1}{16}-\dfrac{1}{144}=0$

$144{{x}^{2}}-144x+36+144{{y}^{2}}-72y+9-1=0$ (Solve by taking LCM)

$144{{x}^{2}}-144x+144{{y}^{2}}-72y+44=0$

$36{{x}^{2}}-36x+36{{y}^{2}}-18y+11=0$

$36{{x}^{2}}+36{{y}^{2}}-36x-18y+11=0$

Ans: The equation of a circle with centre (h,k) and radius (r) is given as

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

It is given that centre (h,k)=(1,1) and radius $(r)=\sqrt{2}.$

Therefore, the equation of the circle is

${{(x-1)}^{2}}+{{(y-1)}^{2}}={{(\sqrt{2})}^{2}}$

$\begin{align}  & {{x}^{2}}-2x+1+{{y}^{2}}-2y+1=2 \\  & {{x}^{2}}+{{y}^{2}}-2x-2y=0 \\ \end{align}$

Ans: The equation of a circle with centre (h,k) and radius (r) is given as

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

It is given that centre (h,k)=(-a,-b) and radius $(r)=\sqrt{{{a}^{2}}-{{b}^{2}}}.$

Therefore, the equation of the circle is

${{(x+a)}^{2}}+{{(y+b)}^{2}}={{(\sqrt{{{a}^{2}}-{{b}^{2}}})}^{2}}$

$\begin{align}   & {{x}^{2}}+2ax+{{a}^{2}}+{{y}^{2}}+2by+{{b}^{2}}={{a}^{2}}-{{b}^{2}} \\  & {{x}^{2}}+{{y}^{2}}+2ax+2by+2{{b}^{2}}=0 \\ \end{align}$

Ans: The equation of the given circle is ${{(x+5)}^{2}}+{{(y-3)}^{2}}=36$

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

$\Rightarrow {{\{x-(-5)\}}^{2}}+{{(y-3)}^{2}}={{(6)}^{2}}$, which is the form of ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$, where $h=-5,k=3,and\; r=6$.

Thus, the centre of the given circle is $(-5,3),$ while its radius is 6.

Ans: The equation of the given circle is ${{x}^{2}}+{{y}^{2}}-4x-8y-45=0$

$\Rightarrow {{x}^{2}}+{{y}^{2}}-4x-8y-45=0$

$\Rightarrow ({{x}^{2}}-4x)+({{y}^{2}}-8y)=45$

$\Rightarrow \{{{x}^{2}}-2(x)(2)+{{(2)}^{2}}\}+\{{{y}^{2}}-2(y)(4)+{{(4)}^{2}}\}-4-16=45$

$\Rightarrow {{(x-2)}^{2}}+{{(y-4)}^{2}}=65$

$\Rightarrow {{(x-2)}^{2}}+{{(y-4)}^{2}}={{\left( \sqrt{65} \right)}^{2}},$ which is of the form ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$, where $h=2,k=4,and\; r=\sqrt{65}$

Thus, the centre of the given circle is $(2,4),$while its radius is $\sqrt{65}$.

Ans: The equation of the given circle is ${{x}^{2}}+{{y}^{2}}-8x+10y-12=0$

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

$\Rightarrow \{{{x}^{2}}-2(x)(4)+{{(4)}^{2}}\}+\{{{y}^{2}}+2(y)(5)+{{(5)}^{2}}-16-25=12$

$\Rightarrow {{(x-4)}^{2}}+{{(y+5)}^{2}}=53$

$\Rightarrow {{(x-4)}^{2}}+{{\{y-(-5)\}}^{2}}={{\left( \sqrt{53} \right)}^{2}},$ which is of the form ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$, where $h=4,k=-5,and\; r=\sqrt{53}$

Thus, the centre of the given circle is $(4,-5),$ while its radius is $\sqrt{53}$.

Ans: The equation of the given circle is $2{{x}^{2}}+2{{y}^{2}}-x=0$

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

$\Rightarrow 2\left[ \left( {{x}^{2}}-\dfrac{x}{2} \right)+{{y}^{2}} \right]=0$

$\Rightarrow \left\{ {{x}^{2}}-2.x\left( \dfrac{1}{4} \right)+{{\left( \dfrac{1}{4} \right)}^{2}} \right\}+{{y}^{2}}-{{\left( \dfrac{1}{4} \right)}^{2}}=0$

$\Rightarrow {{\left( x-\dfrac{1}{4} \right)}^{2}}+{{\left( y-0 \right)}^{2}}={{\left( \dfrac{1}{4} \right)}^{2}},$ which is of the form ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$, where $h=\dfrac{1}{4},k=0,$ and $r = \dfrac{1}{4}$

Thus, the centre of the given circle is $\left( \dfrac{1}{4},0 \right),$ while its radius is $\dfrac{1}{4}$.

Ans: Let the equation of the required circle be ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$

Since the circle passes through points $(4,1)$ and $(6,5)$,

${{(4-h)}^{2}}+{{(1-k)}^{2}}={{r}^{2}}$ ………(i)

${{(6-h)}^{2}}+{{(5-k)}^{2}}={{r}^{2}}$ ………(ii)

Since the centre (h,k) of the circle lies on line $4x+y=16,$

$4h+k=16$ ………(iii)

From equations (i) and (ii), we get

$\Rightarrow {{(4-h)}^{2}}+{{(1-k)}^{2}}={{(6-h)}^{2}}+{{(5-k)}^{2}}$

$\Rightarrow 16-8h+{{h}^{2}}+1-2k+{{k}^{2}}=36-12h+{{h}^{2}}+25-10k+{{k}^{2}}$

$\begin{align}   & \Rightarrow 16-8h+1-2k=36-12h+25-10k \\  & \Rightarrow 4h+8k=44 \\ \end{align}$

$\Rightarrow h+2k=11$ ………(iv)

On solving equations (iii) and (iv), we obtain $h=3$ and $k=4$

On substituting the values of h and k in equation (i), we obtain

${{(4-3)}^{2}}+{{(1-4)}^{2}}={{r}^{2}}$

$\Rightarrow {{(1)}^{2}}+{{(-3)}^{2}}={{r}^{2}}$

$\Rightarrow 1+9={{r}^{2}}$

$\begin{align} & \Rightarrow {{r}^{2}}=10 \\  & \Rightarrow r=\sqrt{10} \\ \end{align}$

Thus, the equation of the required circle is

${{(x-3)}^{2}}+{{(y-4)}^{2}}={{\left( \sqrt{10} \right)}^{2}}$

$\begin{align}   & {{x}^{2}}-6x+9+{{y}^{2}}-8y+16=10 \\  & {{x}^{2}}+{{y}^{2}}-6x-8y+15=0 \\ \end{align}$