- Answer all questions in PART- A. Each question carries one mark.
- Answer any ten questions in PART- B. Each question carries two marks.
- Answer all questions by selecting either A or B. Each question carries fifteen marks. (7 + 8)
Clarks Table and programmable calculators are not permitted.
\[\underline{PART\ -\ A}\]
\[1.\ \color{green}{Find\ the\ combined\ equation\ of\ the\ two\ straight\ lines\ represented\ by}\ \hspace{7cm}\]\[\color{green}{2x\ +\ 3y\ =\ 0\ and\ 4x\ -\ 5y\ =\ 0}\ \hspace{8cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{5cm}\ The\ two\ separate\ lines\ are\ 2x\ +\ 3y\ =0\ and\ 4x\ -\ 5y\ =\ 0\ \hspace{10cm}\]
\[The\ combined\ equation\ is\]
\[\hspace{2cm}\ (2x\ +\ y) (4x\ -\ 5y)\ =\ 0\]
\[\hspace{2cm}\ 6x^2\ -\ 10xy\ +\ 4xy\ -\ 5y^2\ =\ 0\]
\[\hspace{2cm}\ 6x^2\ -\ 6xy\ -\ 5y^2\ =\ 0\]
\[2.\ \color{green}{Find\ the\ Unit\ vector\ parallel\ to}\ \hspace{10cm}\]\[\color{green}{2\overrightarrow{i}\ – \overrightarrow{j}\ +\ 4\overrightarrow{k}}\ \hspace{8cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{a}\ =\ 2\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ 4\overrightarrow{k}\]
\[\overrightarrow{|a|} = \sqrt{(2)^2 + (-1)^2+(4)^2 }\]
\[= \sqrt{(4\ +\ 1\ +\ 16)}\]
\[=\sqrt{21}\]
\[\overrightarrow{|a|}=\sqrt{21}\]
\[Unit\ vector\ along\ \overrightarrow{a}=\frac{\overrightarrow{a}}{\overrightarrow{|a|}}= \frac{3\overrightarrow{i}\ + 4\overrightarrow{j}- 5\overrightarrow{k}}{\sqrt{50}}\]
\[3.\ \color{green}{Find\ the\ value\ of\ [\overrightarrow{i} + \overrightarrow{j}\ \overrightarrow{j} + \overrightarrow{k}\ \overrightarrow{k} + \overrightarrow{i}]}\ \hspace{7cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{a}= \overrightarrow{i}\ + \overrightarrow{j}\]
\[\overrightarrow{b}= \overrightarrow{j}\ + \overrightarrow{k}\]
\[\overrightarrow{c}= \overrightarrow{k} + \overrightarrow{i}\]
\[ [\overrightarrow{a}\ \overrightarrow{b} \overrightarrow{c}] =\begin{vmatrix}
1 & 1 & 0\\
0 & 1 & 1\\
1 & 0 & 1\\
\end{vmatrix}\]
\[=\ 1(1\ -\ 0)\ -\ 1(0\ -\ 1)\ +\ 0(0\ -\ 1)\]
\[=\ 1(1)\ -\ 1(-1)\ +\ 0\]
\[=\ 1\ +\ 1\]
\[=\ 2\]
\[4.\ \color {green}{Evaluate: \int\ (x^3\ -\ x\ -\ 2)\ dx}\ \hspace{15cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int(x^3\ -\ x\ -\ 2)\ dx = \frac{x^4}{4}\ -\ \frac{x^2}{2}\ -\ 2x\ + c\]
\[5.\ \color {green}{Evaluate: \int_1^3 \frac{dx}{x}}\ \hspace{17cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int_1^3 \frac{1}{x}\ dx = log\ x \Biggr]_{1}^{3}\]
\[=[log\ 3 – log\ 1]\]
\[= log\ 3 – 0\]
\[= log\ 3\]
\[\boxed{\int_1^2 \frac{1}{x}\ dx = log\ 3}\]
\[\underline{PART\ -\ B}\]
\[6.\ \color{green}{Find\ the\ equation\ of\ the\ straight\ line\ parallel\ to\ the\ line\ 3x\ +\ 2y\ -\ 7\ =\ 0}\ \hspace{6cm}\]\[\color{green}{and\ passing\ through\ the\ point\ (1, -2)}\ \hspace{8cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[Let\ the\ equation\ of\ line\ parallel\ to\ 3x\ +\ 2y\ -\ 7\ =0\ – – -\ (1)\]
\[is\ 3x\ +\ 2y\ +\ k\ =0\ —-\ (2)\]
\[\text{Equation (2) passes through ( 1, -2 )}\]
\[\text{ put x= 1, y =-2 in equation (2)}\]
\[\text{3(1) + 2(-2) +k = 0}\]
\[\text{3 – 4 + k = 0}\]
\[k = 1\]
\[\text{∴ Required line is 3x + 2y +1 = 0}\]
\[7.\ \color{green}{Find\ the\ centre\ and\ radius\ of\ the\ circle\ x^2\ +\ y^2\ -\ 8\ y\ +\ 3\ =\ 0}\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\color {blue} {Soln:}\ Given\ x^2\ +\ y^2\ -\ 8\ y\ +\ 3\ =\ 0\ \hspace{10cm}\]
\[We\ know\ that\ the\ equation\ of\ circle\ is\ x^2\ +\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\ \hspace{10cm}\]
\[2\ g\ =\ 0\ \hspace{3cm}\ 2\ f\ =\ -\ 8\ \hspace{3cm}\ c\ =\ 3\]
\[g\ =\ 0\ \hspace{3cm}\ f\ =\ -\ 4\ \hspace{3cm}\]
\[centre\ =\ (-\ g,\ -\ f)\ \hspace{4cm}\ r\ =\ \sqrt{(g^2\ +\ f^2\ -\ c)}\]
\[centre\ =\ (0,\ 4)\ \hspace{4cm}\ r\ =\ \sqrt{(0^2\ +\ (-4)^2\ -\ 3)}\]
\[\hspace{6cm}\ r\ =\ \sqrt{(0\ +\ 16\ -\ 3)}\]
\[\hspace{6cm}\ r\ =\ \sqrt{13}\]
\[centre = (0, 4)\ \hspace{5cm}\ r\ =\ \sqrt{13}\]
\[8.\ \color{green}{Show\ that\ the\ equation\ x^2\ +\ 4y^2\ -\ 4\ x\ +\ 24\ y\ +\ 31\ =\ 0}\ \hspace{10cm}\]\[\color {green} {represents\ an\ ellipse}\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ \hspace{18cm}\]
\[\color {blue} {Soln:}\ x^2\ +\ 4y^2\ -\ 4\ x\ +\ 24\ y\ +\ 31\ =\ 0\ ———- (1)\ \hspace{10cm}\]
\[\hspace{2cm}\ Condition\ for\ (1)\ to\ represent\ an\ ellipse\ is\ h^2\ -\ ab\ \lt\ 0\ \hspace{8cm}\]
\[Comparing\ with\ a\ x^2\ +\ 2h\ x\ y\ +\ by^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ = 0\]
\[We\ get\ a\ =\ 1,\ b\ =\ 4\ \hspace{5cm}\ 2h\ =\ 0,\ \implies\ h\ =\ 0\ \hspace{5cm}\]
\[\hspace{2cm}\ h^2\ -\ ab\ =\ (0)^2\ -\ 1(4)\ =\ -\ 4\ \lt\ 0\ \hspace{8cm}\]
\[\therefore\ (1)\ represents\ an\ ellipse\ \hspace{10cm}\]
\[9.\ \color{green}{Find\ the\ projection\ of\ the\ vector}\ \hspace{12cm}\]\[\color{green}{3\overrightarrow{i}+ 4\overrightarrow{j}\ +\ 5\overrightarrow{k} on\ the\ vector\ \overrightarrow{i}+ 2\overrightarrow{j}\ +\ 6\overrightarrow{k}}\ \hspace{8cm}\]
\[\color {blue} {Soln:}\ \hspace{18cm}\]
\[\overrightarrow{a}= 3\overrightarrow{i}+ 4\overrightarrow{j}\ +\ 5\overrightarrow{k} \]
\[\overrightarrow{b}= \overrightarrow{i}\ +\ 2 \overrightarrow{j}\ + 6\overrightarrow{k} \]
\[Projection\ of\ \overrightarrow{a} on \overrightarrow{b} =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|b|}}\]
\[=\frac{(3\overrightarrow{i}\ +\ 4\overrightarrow{j}\ +\ 5\overrightarrow{k}).(\overrightarrow{i}+2\overrightarrow{j}\ +\ 6\overrightarrow{k})}{\sqrt{(1)^2\ +\ (2)^2\ +\ (6)^2 }}\]
\[= \frac{3(1)\ +\ 4(2)\ +\ 5(6)}{\sqrt{(1 + 4 + 36 }}\]
\[= \frac{3\ +\ 8\ +\ 30}{\sqrt{41}}\]
\[\boxed{Projection\ of\ \overrightarrow{a} on \overrightarrow{b} =\frac{41}{\sqrt{41}}}\]
\[10.\ \color{green}{Show\ that\ the\ vectors}\ \hspace{15cm}\]\[\color{green}{4\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 3\overrightarrow{k}\ and\ 3\overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ 5\overrightarrow{k} are\ perpendicular}\ \hspace{8cm}\]
\[\color {blue} {Soln:}\ \hspace{18cm}\]
\[\overrightarrow{a}\ =\ 4\overrightarrow{i}\ +\ \overrightarrow{j}\ – 3\ \overrightarrow{k} \]
\[\overrightarrow{b}= 3\overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ 5\overrightarrow{k} \]
\[ \overrightarrow{a}.\overrightarrow{b}\ =\ (4\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 3\overrightarrow{k}) .(3\overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ 5\overrightarrow{k})\]
\[=\ 4(3)\ +\ 1(3)\ -\ 3(5)\]
\[=\ 12\ +\ 3\ -\ 15\]
\[=\ 0\]
\[\therefore\ \overrightarrow{a}\ and\ \overrightarrow{b}\ are\ perpendicular\ vectors.\]
\[11.\ \color{green}{Find\ \overrightarrow{a}× \overrightarrow{b}}\ \hspace{15cm}\]\[\color{green}{if\ \overrightarrow{a}= \overrightarrow{i}+ \overrightarrow{j}+ \overrightarrow{k} and \overrightarrow{b}= 2\overrightarrow{i}- \overrightarrow{j}\ +\ \overrightarrow{k}}\ \hspace{8cm}\]
\[\color {blue} {Soln:}\ \hspace{18cm}\]
\[\overrightarrow{a}= \overrightarrow{i}+ \overrightarrow{j}+\overrightarrow{k} \]
\[\overrightarrow{b}= 2\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ \overrightarrow{k} \]
\[\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\
1 & 1 & 1\\
2 & -1 & 1\\
\end{vmatrix}\]
\[ = \overrightarrow{i}( 1 + 1)\ -\ \overrightarrow{j}(1\ -\ 2)\ +\ \overrightarrow{k}(-1\ -\ 2)\]
\[ = \overrightarrow{i}(2)\ -\ \overrightarrow{j}(-\ 1)\ +\ \overrightarrow{k}(-3)\]
\[\boxed{ \overrightarrow{a}× \overrightarrow{b}\ =\ 2\overrightarrow{i}\ +\ \overrightarrow{j}\ -3\overrightarrow{k}}\]
\[12.\ \color{green}{Find\ the\ value\ of\ ‘m’\ so\ that\ the\ vectors}\ \hspace{10cm}\]\[\color{green}{2\ \overrightarrow{i}\ – \overrightarrow{j}\ +\ \overrightarrow{k},\ \overrightarrow{i}\ +\ 2\ \overrightarrow{j}\ -\ 3\ \overrightarrow{k}\ ,\ 3\overrightarrow{i}\ +\ m\ \overrightarrow{j}\ +\ 5\ \overrightarrow{k}\ coplanar}\ \hspace{8cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{a}= 2\overrightarrow{i}- \overrightarrow{j}+ \overrightarrow{k}\]
\[\overrightarrow{b}= \overrightarrow{i}+ 2\overrightarrow{j}- 3\overrightarrow{k}\]
\[\overrightarrow{c}= 3\overrightarrow{i}+ m\overrightarrow{j}+ 5\overrightarrow{k}\]
\[Given\ \overrightarrow{a},\overrightarrow{b} and\ \overrightarrow{c} are\ coplanar\ \implies [\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}] = 0\]
\[\begin{vmatrix}
2 &- 1 & 1\\
1 & 2 & -3\\
3 & m & 5\\
\end{vmatrix}=0\]
\[2(10\ +\ 3\ m)\ +\ 1(5\ +\ 9)\ +\ 1(m\ -\ 6)\ =\ 0\]
\[20\ +\ 6m\ +\ 14\ +\ m\ -\ 6\ =\ 0\]
\[28\ +\ 7m\ =\ 0\]
\[7m\ =\ -\ 28\]
\[m\ =\ -\ 4\]
\[13.\ \color{green}{Find\ the\ gradient\ of\ \phi = xyz}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{18cm}\]
\[ \phi = xyz\]
\[\nabla\ \phi=(\frac{\partial\ \phi}{\partial\ x})\overrightarrow{i} + (\frac{\partial\ \phi}{\partial\ y})\overrightarrow{j} + (\frac{\partial\ \phi}{\partial\ z})\overrightarrow{k}————(1)\]
\[\frac{\partial\ \phi}{\partial\ x} = yz \]
\[\frac{\partial\ \phi}{\partial\ y} = xz\]
\[\frac{\partial\ \phi}{\partial\ z} = xy\]
\[Equation\ (1)\ becomes\ \hspace{10cm}\]
\[\nabla\ \phi= yz\overrightarrow{i} + xz\overrightarrow{j} + xy\overrightarrow{k}\]
\[14.\ \color{green}{Find\ div\ \overrightarrow{F}}\ \hspace{18cm}\]\[\color{green}{if\ \overrightarrow{F}\ =\ 3\ x^2\ \overrightarrow{i}\ +\ 5\ x\ y^2\ \overrightarrow{j}\ +\ x\ y\ z^3\ \overrightarrow{k}\ at\ the\ point\ (1,\ 2,\ 3)}\ \hspace{10cm}\]
\[\color {blue} {Soln:}\ \hspace{18cm}\]
\[ \overrightarrow{F}\ =\ 3\ x^2\ \overrightarrow{i}\ +\ 5\ x\ y^2\ \overrightarrow{j}\ +\ x\ y\ z^3\ \overrightarrow{k}\]
\[\nabla\ .\ F\ =\ \frac{\partial\ F}{\partial\ x}\ +\ \frac{\partial\ F}{\partial\ y}\ +\ \frac{\partial\ F}{\partial\ z}\ ————(1)\]
\[\frac{\partial\ F}{\partial\ x}\ =\ 6\ x\ +\ 5\ y^2\ +\ y\ z^3\]
\[\frac{\partial\ F}{\partial\ y}\ =\ 10\ x\ y\ +\ 5\ y^2\ +\ x\ z^3\]
\[\frac{\partial\ F}{\partial\ z}\ =\ 3\ x\ y\ z^2\]
Equation (1) becomes
\[\nabla\ .\ F\ =\ 6\ x\ +\ 5\ y^2\ +\ y\ z^3 \ +\ 10\ x\ y\ +\ 5\ y^2\ +\ x\ z^3\ +\ 3\ x\ y\ z^2\]
At (1, 2, 3)
\[\nabla\ .\ F\ =\ 6\ (1)\ +\ 5\ (2)^2\ +\ 2\ (3)^3 \ +\ 10\ (1)\ (2)\ +\ 5\ (2)^2\ +\ 1\ (3)^3\ +\ 3\ (1)\ (2)\ (3)^2\]
\[=\ 6\ +\ 5\ (4)\ +\ 2\ (9) \ +\ 10\ (2)\ +\ 5\ (4)\ +\ 1\ (3)\ +\ 3\ (18)\]
\[=\ 6\ +\ 20\ +\ 18 \ +\ 20\ +\ 20\ +\ 3\ +\ 54\]
\[\nabla\ .\ F\ =\ 141\]
\[15.\ \color {green}{Evaluate: \int(x + 3 ) ( x + 2) \ dx}\ \hspace{15cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int(x + 3 ) ( x + 2)\ dx\]
\[= \int( x^2 +\ 3x\ +\ 2x\ +\ 6)\ dx\]
\[= \int(x^2\ +\ 5x\ +\ 6)\ dx\]
\[ =\ \frac{x^3}{3}\ +\ 5\frac{x^2}{2}\ +\ 6x + c\]
\[ =\ \frac{x^3}{3}\ +\ \frac{5}{2}\ x^2\ +\ 6x + c\]
\[16.\ \color {green}{Evaluate: \int\ (3 sin x + 9)\ dx}\ \hspace{15cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int(3 sin x\ +\ 9)\ dx\ =\ 3\ \int sin x\ dx\ +\ 9\ \int 1\ dx\]
\[=- 2 cos x\ +\ 9x + c\]
\[\therefore\ \boxed{\int(3 sin x\ +\ 9)\ dx\ =\ – 2\ cos x\ + 9\ x + c}\]
\[17.\ \color {green}{Evaluate: \int\ \frac{2x}{x^2 +\ 4}\ dx}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[put\ u\ =\ x^2\ +\ 4\]
\[\frac{du}{dx}= \ 2x\]
\[du\ =\ 2\ dx\]
\[\int\ \frac{2x}{x^2\ +\ 4}\ dx= \int\ \frac{du}{u}\]
\[=\ log\ u\ +\ c\]
\[= log(x^2\ +\ 4) + c\]
\[\boxed{\int\ \frac{2x}{x^2\ +\ 4}\ dx= log(x^2\ +\ 4) + c}\]
\[18.\ \color {green}{Evaluate: \int\ x^2\ sin\ 2x\ dx}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[ ILATE \]
\[u= x^2\ \hspace{2cm}\ dv = sin\ 2x\ dx\]
\[u^! = 2x\ \hspace{2cm}\ v = – \frac{cos\ 2x}{2}\]
\[u^{!!} = 2\ \hspace{2cm}\ v_1 = – \frac{sin\ 2x}{4}\]
\[\hspace{3cm}\ v_2 = \frac{cos\ 2x}{8}\]
\[\int u\ dv = uv – u^!v_1 + u^{!!}v_2 \]
\[\int x^2\ sin\ 2x\ dx = x^2(-\frac{cos\ 2x}{2}) – 2x (-\frac{sin\ 2x}{4}) + 2(\frac{cos\ 2x}{8}) + c\]
\[ = – x^2\frac{cos\ 2x}{2} + x \frac{sin\ 2x}{2} + \frac{cos\ 2x}{4} + c\]
\[19.\ \color {green}{Evaluate: \int\ log\ x\ dx}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[\int log\ x\ dx = \int 1.\ log\ x\ dx\]
ILATE
Google ad
\[u= log\ x\ \hspace{2cm}\ dv = 1\ dx\]
\[\frac{du}{dx} = \frac{d}{dx} ( log\ x)\ \hspace{2cm}\ \int dv = \int 1\ dx\]
\[\frac{du}{dx} = \frac{1}{x}\ \hspace{2cm}\ v = x\]
\[\ du = \frac{1}{x}\ dx\ \hspace{5cm}\]
\[\int u\ dv = uv – \int v\ du\]
\[\int log\ x\ dx = x\ log\ x- \int x\ (\frac{1}{x})\ dx\]
\[=\ x\ log\ x- \int 1\ dx\]
\[=\ x\ log\ x- x + c\]
\[\boxed{\int log\ x\ dx = x\ log\ x- x +c}\]
\[20.\ \color {green}{Evaluate: \int\ x\ e^{3x}\ dx}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[ ILATE \]
\[u= x\ \hspace{2cm}\ dv = e^{3x}\ dx\]
\[\frac{du}{dx} = \frac{d}{dx} (x)\ \hspace{2cm}\ \int dv = \int e^{3x}\ dx\]
\[\frac{du}{dx} = 1\ \hspace{2cm}\ v = \frac{e^{3x}}{3}\]
\[\ du = dx\ \hspace{5cm}\]
\[\int u\ dv = uv – \int v\ du\]
\[\int x\ e^{3x} dx = x\ \frac{e^{3x}}{3} – \int \frac{e^{3x}}{3}\ dx\]
\[= x\ \frac{e^{3x}}{3} – \frac{e^{3x}}{9}\ + c\]
\[\boxed{\int x\ e^{3x}\ dx = x\ \frac{e^{3x}}{3} – \frac{e^{3x}}{9} +c}\]
\[\underline{PART\ -\ C}\]
\[21.\ A)\ i.\ \color{green}{Find\ the\ equation\ of\ the\ circle\ on\ the\ line\ joining\ the\ points\ (2,3),\ (-\ 4,\ 5)\ as\ diameter.}\ \hspace{5cm}\]\[\color{green}{Aslo\ find\ the\ centre\ and\ radius\ of\ the\ circle}\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ \hspace{18cm}\]
\[\hspace{1cm}\ ii.\ \color{green}{Prove\ that\ the\ circles\ x^2\ +\ y^2\ -\ 4x\ -\ 6y\ +\ 9\ = 0}\ \hspace{5cm}\]\[\color{green}{and\ x^2\ +\ y^2\ +\ 2x\ +\ 2y\ -\ 7\ = 0\ touch\ each\ other.}\ \hspace{5cm}\]\[\color{green}{\text{Find the co-ordinates of the point of contact.}}\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[Given\ x^2 + y^2 -\ 4x\ -\ 6y\ +\ 9\ = 0 ——————— (1)\]
\[Given\ x^2\ +\ y^2\ +\ 2x\ +\ 2y\ -\ 7\ = 0 ——————— (2)\]
\[From\ (1)\ \hspace 10cm\]
\[2g_1 =\ -\ 4\ \hspace 2cm\ 2f_1\ =\ -\ 6\ \hspace 2cm\ c_1 =\ 9\]
\[g_1 =\ -\ 2\ \hspace 2cm\ f_1 =\ -\ 3\ \hspace 2cm\ c_1 =\ 9\]
\[Centre\ is\ C_1 = (-g_1,\ -f_1)\ \hspace 10cm\ r_1 = \sqrt{g_1^2 + f_1^2 -c_1}\]
\[ C_1 = (2,\ \ 3)\ \hspace 10cm\ r_1 = \sqrt{(-2)^2\ +\ (-3)^2\ -\ 9}\]
\[ \hspace 10cm\ r_1 = \sqrt{4\ +\ 9\ -\ 9}\]
\[ \hspace 10cm\ r_1 = \sqrt{4}\ =\ 2\]
\[\boxed{C_1 = ( 2, 3)\ and\ r_1 =\ 2}\]
\[From\ (2)\ \hspace 10cm\]
\[2g_2 =\ 2\ \hspace 2cm\ 2f_2\ =\ 2\ \hspace 2cm\ c_2 =\ -\ 7\]
\[g_2 =\ 1\ \hspace 2cm\ f_2 =\ 1\ \hspace 2cm\ c_2 =\ -\ 7\]
\[Centre\ is\ C_2 = (-g_2,\ -f_2)\ \hspace 10cm\ r_2 = \sqrt{g_2^2 + f_2^2 -c_2}\]
\[ C_2 = (-\ 1\ , -\ 1)\ \hspace 10cm\ r_2 = \sqrt{(1)^2\ +\ (1)^2\ +\ 7}\]
\[ \hspace 10cm\ r_2\ = \sqrt{1\ + 1\ +\ 7}\]
\[ \hspace 10cm\ r_2 = \sqrt{9} =\ 3\]
\[\boxed{C_2 = (-\ 1, -\ 1)\ and\ r_2 =\ 3}\]
\[C_1C_2 = \sqrt{(-1\ -\ 2)^2\ +\ (-1\ -\ 3)^2}\]
\[C_1C_2 = \sqrt{(-3)^2\ +\ (-4)^2}\]
\[C_1C_2 = \sqrt{9\ +\ 16}\]
\[C_1C_2 = \sqrt{25}\ =\ 5\]
\[r_1\ +\ r_2\ =\ 2\ +\ 3\ =\ 5\ =\ C_1C_2\]
\[\boxed{C_1C_2 = r_1 + r_2}\]
\[The\ given\ circles\ touch\ each\ other\ externally\]
To find the point of contact:
For externally touching circles, the point of contact lies on the line segment joining the centers of the two circles. We can find the point of contact using the section formula, dividing the line segment joining C1 and C2 in the ratio of the radii r1:r2.
\[\boxed{C_1 = (2, 3)\ ,\ C_2 =\ (-\ 1, -\ 1)\ and\ r_1\ =\ 2.\ r_2\ =\ 3}\]
\[Here\ x_1\ =\ 2\ ,\ y_1\ =\ 3\ and\ x_2\ =\ -1\ ,\ y_2\ =\ -1\]
\[P(x\ ,\ y)\ =\ (\frac{r_1\ x_2\ +\ r_2\ x_1}{r_1\ +\ r_2},\ \frac{r_1\ y_2\ +\ r_2\ y_1}{r_1\ +\ r_2})\]
\[=\ (\frac{2(-1)\ +\ 3(2)}{2\ +\ 3},\ \frac{2(-1)\ +\ 3(3)}{2\ +\ 3})\]
\[=\ (\frac{-2\ +\ 6}{5},\ \frac{-2\ +\ 9}{5})\]
\[P(x\ ,\ y)\ =\ (\frac{4}{5},\ \frac{7}{5})\]
Google ad
Google ad
