NCERT Class 11 Maths Conic Sections

NCERT Class 11 Maths Conic Sections

Chapter 11 Ex.11.1 Question 1

Find the equation of the circle with centre \(\left( {0,\;2} \right)\) and radius \(2.\)

Solution

The equation of a circle with centre\(\left( {h,\;k} \right)\) and radius \(r\) is given as

\[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\]

It is given that centre \(\left( {h,\;k} \right) = \left( {0,\;2} \right)\) and radius \(\left( r \right) = 2\).

Therefore, the equation of the circle is

\[\begin{align}{\left(x - 0 \right)^2} + {\left( y - 2 \right)^2} &= {2^2}\\{x^2} + {y^2} - 4y + 4 &= 4\\{x^2} + {y^2} - 4y &= 0\end{align}\]

Chapter 11 Ex.11.1 Question 2

Find the equation of the circle with centre \(\left( { - 2,\;3} \right)\) and radius \(4.\)

Solution

The equation of a circle with centre \(\left( {h,\;k} \right)\) and radius \(r\) is given as

\[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\]

It is given that centre \(\left( {h,\;k} \right) = \left( { - 2,\;3} \right)\)and radius \(\left( r \right) = 4\).

Therefore, the equation of the circle is

\[\begin{align}{\left( {x + 2} \right)^2} + {\left( {y - 3} \right)^2} &= {4^2}\\{x^2} + 4x + 4 + {y^2} - 6y + 9 &= 16\\{x^2} + {y^2} + 4x - 6y - 3 &= 0\end{align}\]

Chapter 11 Ex.11.1 Question 3

Find the equation of the circle with centre \(\left( {\frac{1}{2},\;\frac{1}{4}} \right)\) and radius \(\left( {\frac{1}{{12}}} \right)\)

Solution

The equation of a circle with centre \(\left( {h,\;k} \right)\) and radius \(r\) is given as

\[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\]

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

Therefore, the equation of the circle is

\[\begin{align}{\left( {x - \frac{1}{2}} \right)^2} + {\left( {y - \frac{1}{4}} \right)^2} &= {\left( {\frac{1}{{12}}} \right)^2}\\{x^2} - x + \frac{1}{4} + {y^2} - \frac{y}{2} + \frac{1}{{16}} & = \frac{1}{{144}}\\{x^2} - x + \frac{1}{4} + {y^2} - \frac{y}{2} + \frac{1}{{16}} - \frac{1}{{144}} &= 0\\144{x^2} - 144x + 36 + 144{y^2} - 72y + 9 - 1 &= 0\\144{x^2} - 144x + 144{y^2} - 72y + 44 &= 0\\36{x^2} - 36x + 36{y^2} - 18y + 11 &= 0\\36{x^2} + 36{y^2} - 36x - 18y + 11 &= 0\end{align}\]

Chapter 11 Ex.11.1 Question 4

Find the equation of the circle with centre \(\left( {1,\;1} \right)\) and radius \(\sqrt 2 \)

Solution

The equation of a circle with centre \(\left( {h,\;k} \right)\) and radius \(r\) is given as

\[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\]

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

Therefore, the equation of the circle is

\[\begin{align}{\left( {x - 1} \right)^2} + {\left( {y - 1} \right)^2} &= {\left( {\sqrt 2 } \right)^2}\\{x^2} - 2x + 1 + {y^2} - 2y + 1 &= 2\\{x^2} + {y^2} - 2x - 2y &= 0\end{align}\]

Chapter 11 Ex.11.1 Question 5

Find the equation of the circle with centre \(\left( { - a,\; - b} \right)\) and radius \(\sqrt {{a^2} - {b^2}} \)

Solution

The equation of a circle with centre \(\left( {h,\;k} \right)\) and radius \(r\) is given as

\[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\]

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

Therefore, the equation of the circle is

\[\begin{align}{\left( {x + a} \right)^2} + {\left( {y + b} \right)^2} &= {\left( {\sqrt {{a^2} - {b^2}} } \right)^2}\\{x^2} + 2ax + {a^2} + {y^2} + 2by + {b^2} &= {a^2} - {b^2}\\{x^2} + {y^2} + 2ax + 2by + 2{b^2} &= 0\end{align}\]

Chapter 11 Ex.11.1 Question 6

Find the centre and radius of the circle \({\left( {x + 5} \right)^2} + {\left( {y - 3} \right)^2} = 36\)

Solution

The equation of the given circle is \({\left( {x + 5} \right)^2} + {\left( {y - 3} \right)^2} = 36\)

\[\begin{align}&\Rightarrow\; {\left( {x + 5} \right)^2} + {\left( {y - 3} \right)^2} = 36\\&\Rightarrow\; {\left[ {x - \left( { - 5} \right)} \right]^2} + {\left( {y - 3} \right)^2} = {6^2}\end{align}\]

which is of the form \({\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\)

Therefore, on comparing both equations we get

\(h = - 5,\;k = 3\) and \(r=6\)

Thus, the centre of the given circle is while its radius is \(6.\)

Chapter 11 Ex.11.1 Question 7

Find the centre and radius of the circle \({x^2} + {y^2} - 4x - 8y - 45 = 0\)

Solution

The equation of the given circle is \({x^2} + {y^2} - 4x - 8y - 45 = 0\).

\[\begin{align}&\Rightarrow \;{x^2} + {y^2} - 4x - 8y - 45 = 0\\&\Rightarrow \;\left( {{x^2} - 4x} \right) + \left( {{y^2} - 8y} \right) = 45\\&\Rightarrow \;\left\{ {{x^2} - 2\left( x \right)\left( 2 \right) + {2^2}} \right\} + \left\{ {{y^2} - 2\left( y \right)\left( 4 \right) + {4^2}} \right\}\\&\quad- 4 - 16 = 45\\&\Rightarrow \;{\left( {x - 2} \right)^2} + {\left( {y - 4} \right)^2} = 65\\&\Rightarrow \;{\left( {x - 2} \right)^2} + {\left( {y - 4} \right)^2} = {\left( {\sqrt {65} } \right)^2}\end{align}\]

which is of the form \({\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\)

Therefore, on comparing both equations we get

\(h = 2,\;k = 4\) and \(r = \sqrt {65} \)\(\sqrt {65} \)

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

Chapter 11 Ex.11.1 Question 8

Find the centre and radius of the circle \({x^2} + {y^2} - 8x + 10y - 12 = 0\)

Solution

The equation of the given circle is \({x^2} + {y^2} - 8x + 10y - 12 = 0\).

\[\begin{align}&\Rightarrow \;{x^2} + {y^2} - 8x + 10y - 12 = 0\\&\Rightarrow \; \left( {{x^2} - 8x} \right) + \left( {{y^2} + 10y} \right) = 12\\&\Rightarrow \; \left\{ {{x^2} - 2\left( x \right)\left( 4 \right) + {4^2}} \right\} + \left\{ {{y^2} + 2\left( y \right)\left( 5 \right) + {5^2}} \right\} \\&\quad- 16 - 25 = 12\\&\Rightarrow \;{\left( {x - 4} \right)^2} + {\left( {y + 5} \right)^2} = 53\\&\Rightarrow \;{\left( {x - 4} \right)^2} + {\left[ {y - \left( { - 5} \right)} \right]^2} = {\left( {\sqrt {53} } \right)^2}\end{align}\]

which is of the form \({\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\)

Therefore, on comparing both equations we get

\(h = 4,\;k = - 5\) and \(r = \sqrt {53} \)

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

Chapter 11 Ex.11.1 Question 9

Find the centre and radius of the circle \(2{x^2} + 2{y^2} - x = 0\)

Solution

The equation of the given circle is \(2{x^2} + 2{y^2} - x = 0\)

\[\begin{align}&\Rightarrow \;2{x^2} + 2{y^2} - x = 0\\&\Rightarrow \;\left( {2{x^2} - x} \right) + 2{y^2} = 0\\&\Rightarrow \;2\left[ {\left( {{x^2} - \frac{x}{2}} \right) + {y^2}} \right] = 0\\&\Rightarrow \;\left\{ {{x^2} - 2\left( x \right)\left( {\frac{1}{4}} \right) + {{\left( {\frac{1}{4}} \right)}^2}} \right\} + {y^2} - {\left( {\frac{1}{4}} \right)^2} = 0\\&\Rightarrow \;{\left( {x - \frac{1}{4}} \right)^2} + {\left( {y - 0} \right)^2} = {\left( {\frac{1}{4}} \right)^2}\end{align}\]

which is of the form \({\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\)

Therefore, on comparing both equations we get

\(h = \frac{1}{4},\;k = 0\) and \(r = \frac{1}{4}\)

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

Chapter 11 Ex.11.1 Question 10

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

Solution

Let the equation of the required circle be \({\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\).

Since the circle passes through the points \(\left( {4,1} \right)\) and \(\left( {6,5} \right)\)

\[\begin{align}{\left( {4 - h} \right)^2} + {\left( {1 - k} \right)^2} &= {r^2} \qquad \ldots \left( 1 \right)\\{\left( {6 - h} \right)^2} + {\left( {5 - k} \right)^2} &= {r^2} \qquad \ldots \left( 2 \right)\end{align}\]

Since the centre \(\left( {h,\;k} \right)\) of the circle lies on the line \(4x + y = 16\)

\[4h + k = 16 \quad \ldots \left( 3 \right)\]

From equations \(\left( 1 \right)\) and \(\left( 2 \right)\), we obtain

\[\begin{align}&\Rightarrow \;{\left( {4 - h} \right)^2} + {\left( {1 - k} \right)^2} = {\left( {6 - h} \right)^2} + {\left( {5 - k} \right)^2}\\&\Rightarrow \;16 - 8h + {h^2} + 1 - 2k + {k^2} = 36 - 12h + {h^2} + 25 - 10k + {k^2}\\&\Rightarrow \;16 - 8h + 1 - 2k = 36 - 12h + 25 - 10k\\&\Rightarrow \;4h + 8k = 44\\&\Rightarrow\; h + 2k = 11 \qquad \qquad \ldots \left( 4 \right)\end{align}\]

On solving equations \(\left( 3 \right)\) and \(\left( 4 \right)\), we obtain

\(h = 3\) and \(k = 4\)

On substituting the values of \(h\) and \(k\) in equation \(\left( 1 \right)\), we obtain

\[\begin{align}&{\left( {4 - 3} \right)^2} + {\left( {1 - 4} \right)^2} = {r^2}\\&\Rightarrow\; {1^2} + {\left( { - 3} \right)^2} = {r^2}\\&\Rightarrow \;1 + 9 = {r^2}\\&\Rightarrow \;{r^2} = 10\\&\Rightarrow \;r = \sqrt {10}\end{align}\]

Thus, the equation of the required circle is

\[\begin{align}{\left( {x - 3} \right)^2} + {\left( {y - 4} \right)^2} &= {\left( {\sqrt {10} } \right)^2}\\{x^2} - 6x + 9 + {y^2} - 8y + 16 &= 10\\{x^2} + {y^2} - 6x - 8y + 15& = 0\end{align}\]

Chapter 11 Ex.11.1 Question 11

Find the equation of the circle passing through the points \(\left( {2,3} \right)\) and \(\left( { - 1,1} \right)\) and whose centre is on the line \(x - 3y - 11 = 0\).

Solution

Let the equation of the required circle be \({\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\).

Since the circle passes through the points \(\left( {2,3} \right)\) and \(\left( { - 1,1} \right)\)

\[\begin{align}{\left( {2 - h} \right)^2} + {\left( {3 - k} \right)^2} = {r^2} \qquad \ldots \left( 1 \right)\\{\left( { - 1 - h} \right)^2} + {\left( {1 - k} \right)^2} = {r^2} \qquad \ldots \left( 2 \right)\end{align}\]

Since the centre \(\left( {h,\;k} \right)\) of the circle passes lies on the line \(x - 3y - 11 = 0\),

\[h - 3k = 11 \qquad \ldots \left( 3 \right)\]

From equations \(\left( 1 \right)\) and \(\left( 2 \right)\), we obtain

\[\begin{align}&{\left( {2 - h} \right)^2} + {\left( {3 - k} \right)^2} = {\left( { - 1 - h} \right)^2} + {\left( {1 - k} \right)^2}\\&\Rightarrow \;4 - 4h + {h^2} + 9 - 6k + {k^2} = 1 + 2h + {h^2} + 1 + {k^2} - 2k\\&\Rightarrow\; 4 - 4h + 9 - 6k = 1 + 2h + 1 - 2k\\&\Rightarrow \;6h + 4k = 11 \qquad \qquad  \ldots \left( 4 \right)\end{align}\]

On solving equations \(\left( 3 \right)\) and \(\left( 4 \right)\), we obtain

\(h = \frac{7}{2}\) and \(k = \frac{{ - 5}}{2}\)

On substituting the values of \(h\) and \(k\) in equation \(\left( 1 \right)\), we obtain

\[\begin{align}&\Rightarrow\; {\left( {2 - \frac{7}{2}} \right)^2} + {\left( {3 + \frac{5}{2}} \right)^2} = {r^2}\\&\Rightarrow\; {\left( {\frac{{4 - 7}}{2}} \right)^2} + {\left( {\frac{{6 + 5}}{2}} \right)^2} = {r^2}\\&\Rightarrow\; {\left( {\frac{{ - 3}}{2}} \right)^2} + {\left( {\frac{{11}}{2}} \right)^2} = {r^2}\\&\Rightarrow\; \frac{9}{4} + \frac{{121}}{4} = {r^2}\\&\Rightarrow\; \frac{{130}}{4} = {r^2}\end{align}\]

Thus, the equation of the required circle is

\[\begin{align}{\left( {x - \frac{7}{2}} \right)^2} + {\left( {y + \frac{5}{2}} \right)^2} &= \frac{{130}}{4}\\{\left( {\frac{{2x - 7}}{2}} \right)^2} + {\left( {\frac{{2y + 5}}{2}} \right)^2} &= \frac{{130}}{4}\\4{x^2} - 28x + 49 + 4{y^2} + 20y + 25 &= 130\\4{x^2} + 4{y^2} - 28x + 20y - 56 &= 0\\4\left( {{x^2} + {y^2} - 7x + 5y - 14} \right) &= 0\\{x^2} + {y^2} - 7x + 5y - 14 &= 0\end{align}\]

Chapter 11 Ex.11.1 Question 12

Find the equation of the circle with radius \(5\) whose centre lies on \(x\)-axis and passes through the point \(\left( {2,\;3} \right)\).

Solution

Let the equation of the required circle be \({\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\).

Since the radius of the circle is \(5\) and its centre lies on the \(x\)-axis, \(k = 0\) and \(r = 5\).

Now, the equation of the circle becomes\({\left( {x - h} \right)^2} + {y^2} = 25\).

It is given that the circle passes through the point \(\left( {2,3} \right)\).

Therefore,

\[\begin{align}&\Rightarrow\; {\left( {2 - h} \right)^2} + {3^2} = 25\\&\Rightarrow \;{\left( {2 - h} \right)^2} = 25 - 9\\&\Rightarrow \;{\left( {2 - h} \right)^2} = 16\\&\Rightarrow \;2 - h = \pm \sqrt {16} \\&\Rightarrow\; 2 - h = \pm 4\end{align}\]

If, \(2 - h = 4\), then \(h = - 2\)

If \(2 - h = - 4\), then \(h = 6\)

When \(h = - 2\), the equation of the circle becomes

\[\begin{align}{\left( {x + 2} \right)^2} + {y^2} &= 25\\{x^2} + 4x + 4 + {y^2} &= 25\\{x^2} + {y^2} + 4x - 21 &= 0\end{align}\]

When \(h = 6\), the equation of the circle becomes

\[\begin{align}{\left( {x - 6} \right)^2} + {y^2} &= 25\\{x^2} - 12x + 36 + {y^2} &= 25\\{x^2} + {y^2} - 12x + 11 & = 0\end{align}\]

Chapter 11 Ex.11.1 Question 13

Find the equation of the circle passing through \(\left( {0,0} \right)\) and making intercepts \(a\) and \(b\) on the coordinate axes.

Solution

Let the equation of the required circle be \({\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\).

Since the circle passes through \(\left( {0,0} \right)\),

\[\begin{align}&{\left( {0 - h} \right)^2} + {\left( {0 - k} \right)^2} = {r^2}\\&\Rightarrow\; {h^2} + {k^2} = {r^2}\end{align}\]

The equation of the circle now becomes \({\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {h^2} + {k^2}\).

It is given that the circle makes intercepts \(a\) and \(b\) on the coordinate axes. This means that the circle passes through points \(\left( {a,0} \right)\) and \(\left( {0,b} \right)\).

Therefore,

\[\begin{align}{\left( {a - h} \right)^2} + {\left( {0 - k} \right)^2} &= {h^2} + {k^2} \qquad \ldots \left( 1 \right)\\{\left( {0 - h} \right)^2} + {\left( {b - k} \right)^2} &= {h^2} + {k^2} \qquad \ldots \left( 2 \right)\end{align}\]

From equation \(\left( 1 \right)\), we obtain

\[\begin{align}&\Rightarrow \;{a^2} + {h^2} - 2ah + {k^2} = {h^2} + {k^2}\\&\Rightarrow \;{a^2} - 2ah = 0\\&\Rightarrow \;a\left( {a - 2h} \right) = 0\\&\Rightarrow\; a = 0{\text{ or }}\left( {a - 2h} \right) = 0\end{align}\]

However, \(a \ne 0\);

Hence,

\[\begin{align}&\left( {a - 2h} \right) = 0\\&\Rightarrow\;h = \frac{a}{2}\end{align}\]

From equation \(\left( 2 \right)\), we obtain

\[\begin{align}&{h^2} + {b^2} - 2bk + {k^2} = {h^2} + {k^2}\\&\Rightarrow \;{b^2} - 2bk = 0\\&\Rightarrow \;b\left( {b - 2k} \right) = 0\\&\Rightarrow \; b = 0\;{\rm{or}}\;\left( {b - 2k} \right) = 0\end{align}\]

However, \(b \ne 0\);

Hence,

\[\begin{align}&\left( {b - 2k} \right) = 0\\&\Rightarrow \; k = \frac{b}{2}\end{align}\]

Thus, the equation of the required circle is

\[\begin{align}&\Rightarrow\; {\left( {x - \frac{a}{2}} \right)^2} + {\left( {y - \frac{b}{2}} \right)^2} = {\left( {\frac{a}{2}} \right)^2} + {\left( {\frac{b}{2}} \right)^2}\\&\Rightarrow \;{\left( {\frac{{2x - a}}{2}} \right)^2} + {\left( {\frac{{2y - b}}{2}} \right)^2} = \frac{{{a^2} + {b^2}}}{4}\\&\Rightarrow\; 4{x^2} - 4ax + {a^2} + 4{y^2} - 4by + {b^2} = {a^2} + {b^2}\\&\Rightarrow \;4{x^2} + 4{y^2} - 4ax - 4by = 0\\&\Rightarrow \;4\left( {{x^2} + {y^2} - ax - by} \right) = 0\\&\Rightarrow \;{x^2} + {y^2} - ax - by = 0\end{align}\]

Chapter 11 Ex.11.1 Question 14

Find the equation of a circle with centre \(\left( {2,2} \right)\) and passes through the point \(\left( {4,5} \right)\).

Solution

The centre of the circle is given as \(\left( {h,k} \right) = \left( {2,2} \right)\)

Since the circle passes through the point \(\left( {4,5} \right)\), the radius \(\left( r \right)\) of the circle is the distance between the points \(\left( {2,2} \right)\)and \(\left( {4,5} \right)\).

Therefore,

\[\begin{align}r &= \sqrt {{{\left( {2 - 4} \right)}^2} + {{\left( {2 - 5} \right)}^2}} \\&= \sqrt {{{\left( { - 2} \right)}^2} + {{\left( { - 3} \right)}^2}} \\&= \sqrt {4 + 9} \\&= \sqrt {13}\end{align}\]

Thus, the equation of the circle is

\[\begin{align}&{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\\&{\left( {x - 2} \right)^2} + {\left( {y - 2} \right)^2} = {\left( {\sqrt {13} } \right)^2}\\&{x^2} - 4x + 4 + {y^2} - 4y + 4 = 13\\&{x^2} + {y^2} - 4x - 4y - 5 = 0\end{align}\]

Chapter 11 Ex.11.1 Question 15

Does the point \(\left( { - 2.5,\;3.5} \right)\) lie inside, outside or on the circle \({x^2} + {y^2} = 25\)?

Solution

The equation of the given circle is \({x^2} + {y^2} = 25\)

\[\begin{align}&\Rightarrow \;{x^2} + {y^2} = 25\\&\Rightarrow \;{\left( {x - 0} \right)^2} + {\left( {y - 0} \right)^2} = {5^2}\end{align}\]

which is of the form \({\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\), where \(h = 0,\;k = 0\) and \(r = 5\)

Therefore, Centre\( \Rightarrow \left( {0,0} \right)\) and radius \( \Rightarrow 5\)

Distance between point \(\left( { - 2.5,\;3.5} \right)\) and centre \(\left( {0,0} \right)\)

\[\begin{align}&= \sqrt {{{\left( { - 2.5 - 0} \right)}^2} + {{\left( {3.5 - 0} \right)}^2}} \\&= \sqrt {6.25 + 12.25} \\&= \sqrt {18.25} \\&= 4.272\end{align}\]

Here, \(4.272 < 5\), this means less than the radius.

Since the distance between the point \(\left( { - 2.5,\;3.5} \right)\) and centre \(\left( {0,0} \right)\) of the circle is less than the radius of the circle, hence, the point \(\left( { - 2.5,\;3.5} \right)\) lies inside the circle.

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