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if-the-circle-x-2-y-2-2y-8-0-and-x-2-y-2-24x-hy-0-cut-orthogonally-determine-the-value-of-h-

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Question Number 37591 by mondodotto@gmail.com last updated on 15/Jun/18
if the circle x^2 +y^2 −2y−8=0 and x^2 +y^2 −24x+hy=0 cut orthogonally, determine the value of h.
$$\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{circle}}\:\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{y}}−\mathrm{8}=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{and}}\:\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} −\mathrm{24}\boldsymbol{{x}}+\boldsymbol{{hy}}=\mathrm{0}\:\boldsymbol{\mathrm{cut}}\:\boldsymbol{\mathrm{orthogonally}}, \\ $$$$\boldsymbol{\mathrm{determine}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{{h}}. \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 15/Jun/18
x^2 +y^2 −2y−8=0 comparing with x^2 +y^2 +2gx+2fy+c=0 centre(−g,−f) radius=(√(g^2 +f^2 −c)) o_1 (0,1) r_1 =(√(0^2 +(−1)^2 −(−8))) =(√9) =3 o_2 (12,−(h/2)) r_2 =(√((−12)^2 +((h/2))^2 )) r_2 =(√(144+(h^2 /4))) o_1 o_2 =(√((12−0)^2 +(−(h/2)−1)^2 )) =(√(144+(h^2 /4)+h+1)) =(√((h^2 /4)+h+145)) r_1 ^2 +r_2 ^2 =(o_1 o_2 )^2 9+144+(h^2 /4)=(h^2 /4)+h+145 h=8 or mdthod clndition for orthogonslity 2g_1 g_2 +2f_1 f_2 =c_1 +c_2 2×0×(−12)+2×(−1)×(h/2)=−8+0 −h=−8 h=8
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{y}−\mathrm{8}=\mathrm{0} \\ $$$${comparing}\:{with}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{gx}+\mathrm{2}{fy}+{c}=\mathrm{0} \\ $$$${centre}\left(−{g},−{f}\right)\:\:{radius}=\sqrt{{g}^{\mathrm{2}} +{f}^{\mathrm{2}} −{c}} \\ $$$${o}_{\mathrm{1}} \left(\mathrm{0},\mathrm{1}\right)\:\:{r}_{\mathrm{1}} =\sqrt{\mathrm{0}^{\mathrm{2}} +\left(−\mathrm{1}\right)^{\mathrm{2}} −\left(−\mathrm{8}\right)}\:=\sqrt{\mathrm{9}}\:=\mathrm{3} \\ $$$${o}_{\mathrm{2}} \left(\mathrm{12},−\frac{{h}}{\mathrm{2}}\right)\:\:{r}_{\mathrm{2}} =\sqrt{\left(−\mathrm{12}\right)^{\mathrm{2}} +\left(\frac{{h}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$${r}_{\mathrm{2}} =\sqrt{\mathrm{144}+\frac{{h}^{\mathrm{2}} }{\mathrm{4}}}\: \\ $$$${o}_{\mathrm{1}} {o}_{\mathrm{2}} =\sqrt{\left(\mathrm{12}−\mathrm{0}\right)^{\mathrm{2}} +\left(−\frac{{h}}{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} }\:\: \\ $$$$=\sqrt{\mathrm{144}+\frac{{h}^{\mathrm{2}} }{\mathrm{4}}+{h}+\mathrm{1}}\:\:=\sqrt{\frac{{h}^{\mathrm{2}} }{\mathrm{4}}+{h}+\mathrm{145}}\: \\ $$$${r}_{\mathrm{1}} ^{\mathrm{2}} +{r}_{\mathrm{2}} ^{\mathrm{2}} =\left({o}_{\mathrm{1}} {o}_{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\mathrm{9}+\mathrm{144}+\frac{{h}^{\mathrm{2}} }{\mathrm{4}}=\frac{{h}^{\mathrm{2}} }{\mathrm{4}}+{h}+\mathrm{145} \\ $$$${h}=\mathrm{8} \\ $$$${or}\:{mdthod}\: \\ $$$${clndition}\:{for}\:{orthogonslity} \\ $$$$\mathrm{2}{g}_{\mathrm{1}} {g}_{\mathrm{2}} +\mathrm{2}{f}_{\mathrm{1}} {f}_{\mathrm{2}} ={c}_{\mathrm{1}} +{c}_{\mathrm{2}} \\ $$$$\mathrm{2}×\mathrm{0}×\left(−\mathrm{12}\right)+\mathrm{2}×\left(−\mathrm{1}\right)×\frac{{h}}{\mathrm{2}}=−\mathrm{8}+\mathrm{0} \\ $$$$−{h}=−\mathrm{8} \\ $$$${h}=\mathrm{8} \\ $$$$ \\ $$

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