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Question Number 22554 by *D¬ B£$T* last updated on 20/Oct/17
find the eqution of normal to  the circle  2x^2 +2y^2 −3x+4y−32=0 at(2,3)
$${find}\:{the}\:{eqution}\:{of}\:{normal}\:{to} \\ $$$${the}\:{circle} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}{y}−\mathrm{32}=\mathrm{0}\:{at}\left(\mathrm{2},\mathrm{3}\right) \\ $$
Answered by ajfour last updated on 20/Oct/17
2x^2 +2y^2 −3x+4y−32=0  ⇒ x^2 −(3/2)x+y^2 +2y−16=0  (x−(3/4))^2 +(y+1)^2 =16+(9/(16))+1  knowing, equation of circle with  centre (x_0 , y_0 ) is   (x−x_0 )^2 +(y−y_0 )^2 =r^2   we see that  x_0 =(3/4)  , y_0 =−1  y−3=((3−(−1))/((2−(3/4)))) (x−2)  y−3=((16)/5)(x−2)  5y−15=16x−32  5y=16x−17 .
$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}{y}−\mathrm{32}=\mathrm{0} \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{2}}{x}+{y}^{\mathrm{2}} +\mathrm{2}{y}−\mathrm{16}=\mathrm{0} \\ $$$$\left({x}−\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} +\left({y}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{16}+\frac{\mathrm{9}}{\mathrm{16}}+\mathrm{1} \\ $$$${knowing},\:{equation}\:{of}\:{circle}\:{with} \\ $$$${centre}\:\left({x}_{\mathrm{0}} ,\:{y}_{\mathrm{0}} \right)\:{is}\: \\ $$$$\left({x}−{x}_{\mathrm{0}} \right)^{\mathrm{2}} +\left({y}−{y}_{\mathrm{0}} \right)^{\mathrm{2}} ={r}^{\mathrm{2}} \:\:{we}\:{see}\:{that} \\ $$$${x}_{\mathrm{0}} =\frac{\mathrm{3}}{\mathrm{4}}\:\:,\:{y}_{\mathrm{0}} =−\mathrm{1} \\ $$$${y}−\mathrm{3}=\frac{\mathrm{3}−\left(−\mathrm{1}\right)}{\left(\mathrm{2}−\frac{\mathrm{3}}{\mathrm{4}}\right)}\:\left({x}−\mathrm{2}\right) \\ $$$${y}−\mathrm{3}=\frac{\mathrm{16}}{\mathrm{5}}\left({x}−\mathrm{2}\right) \\ $$$$\mathrm{5}{y}−\mathrm{15}=\mathrm{16}{x}−\mathrm{32} \\ $$$$\mathrm{5}\boldsymbol{{y}}=\mathrm{16}\boldsymbol{{x}}−\mathrm{17}\:. \\ $$
Commented by *D¬ B£$T* last updated on 20/Oct/17
wow!! sir i love this your method
$${wow}!!\:{sir}\:{i}\:{love}\:{this}\:{your}\:{method} \\ $$
Commented by *D¬ B£$T* last updated on 20/Oct/17
can u plsresolve it? but this time  throw more light especially how  you got x_0  and y. thanks
$${can}\:{u}\:{plsresolve}\:{it}?\:{but}\:{this}\:{time} \\ $$$${throw}\:{more}\:{light}\:{especially}\:{how} \\ $$$${you}\:{got}\:{x}_{\mathrm{0}} \:{and}\:{y}.\:{thanks} \\ $$
Commented by *D¬ B£$T* last updated on 20/Oct/17
thanks sir
$${thanks}\:{sir} \\ $$

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