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Question Number 37252 by Tinkutara last updated on 11/Jun/18

Commented by rahul 19 last updated on 11/Jun/18

3×4=b_1 ×b_2

$$\mathrm{3}×\mathrm{4}=\mathrm{b}_{\mathrm{1}} ×\mathrm{b}_{\mathrm{2}} \\ $$

Answered by ajfour last updated on 11/Jun/18

x_1 =−(5/3) , x_2 =−(5/2) h=((x_1 +x_2 )/2)=−((25)/(12)) y_1 =−(5/b_1 ) , y_2 =−((10)/b_2 ) ....(a) eq. of circle is (x−h)^2 +(y−k)^2 =R^2 if x=0 (y−k)^2 =R^2 −h^2 y_1 y_2 =h^2 +k^2 −R^2 ....(i) y_1 +y_2 =2k (h+(5/3))^2 +k^2 =R^2 ⇒ R^2 −h^2 −k^2 =((10h)/3)+((25)/9) =−((25)/(12))×((10)/3)+((25)/9) =−((75)/(18)) = −((25)/6) from (a) and (i) y_1 y_2 = ((50)/(b_1 b_2 ))=((25)/6) ⇒ b_1 b_2 = 12 .

$${x}_{\mathrm{1}} =−\frac{\mathrm{5}}{\mathrm{3}}\:,\:\:{x}_{\mathrm{2}} =−\frac{\mathrm{5}}{\mathrm{2}} \\ $$$${h}=\frac{{x}_{\mathrm{1}} +{x}_{\mathrm{2}} }{\mathrm{2}}=−\frac{\mathrm{25}}{\mathrm{12}} \\ $$$${y}_{\mathrm{1}} =−\frac{\mathrm{5}}{{b}_{\mathrm{1}} }\:\:,\:\:{y}_{\mathrm{2}} =−\frac{\mathrm{10}}{{b}_{\mathrm{2}} }\:\:\:\:\:\:....\left({a}\right) \\ $$$${eq}.\:{of}\:{circle}\:{is} \\ $$$$\left({x}−{h}\right)^{\mathrm{2}} +\left({y}−{k}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$${if}\:{x}=\mathrm{0} \\ $$$$\left({y}−{k}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} −{h}^{\mathrm{2}} \\ $$$${y}_{\mathrm{1}} {y}_{\mathrm{2}} ={h}^{\mathrm{2}} +{k}^{\mathrm{2}} −{R}^{\mathrm{2}} \:\:\:\:....\left({i}\right) \\ $$$${y}_{\mathrm{1}} +{y}_{\mathrm{2}} =\mathrm{2}{k} \\ $$$$\left({h}+\frac{\mathrm{5}}{\mathrm{3}}\right)^{\mathrm{2}} +{k}^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:{R}^{\mathrm{2}} −{h}^{\mathrm{2}} −{k}^{\mathrm{2}} =\frac{\mathrm{10}{h}}{\mathrm{3}}+\frac{\mathrm{25}}{\mathrm{9}} \\ $$$$\:\:\:\:\:\:=−\frac{\mathrm{25}}{\mathrm{12}}×\frac{\mathrm{10}}{\mathrm{3}}+\frac{\mathrm{25}}{\mathrm{9}}\:=−\frac{\mathrm{75}}{\mathrm{18}}\:=\:−\frac{\mathrm{25}}{\mathrm{6}} \\ $$$${from}\:\left({a}\right)\:{and}\:\left({i}\right) \\ $$$$\:\:\:\:{y}_{\mathrm{1}} {y}_{\mathrm{2}} =\:\frac{\mathrm{50}}{{b}_{\mathrm{1}} {b}_{\mathrm{2}} }=\frac{\mathrm{25}}{\mathrm{6}} \\ $$$$\Rightarrow\:\:\:\:\:\boldsymbol{{b}}_{\mathrm{1}} \boldsymbol{{b}}_{\mathrm{2}} =\:\mathrm{12}\:. \\ $$

Commented by Tinkutara last updated on 11/Jun/18

Thank you very much Sir! I got the answer. ��������

Commented by ajfour last updated on 11/Jun/18

Let eq. of the circle be (x−h)^2 +(y−k)^2 =R^2 for intercepts on x-axis y=0 ⇒ (x−h)^2 +k^2 =R^2 ⇒ x_1 x_2 =h^2 +k^2 −R^2 for intercepts on y-axis x=0 ⇒ (y−k)^2 +h^2 =R^2 ⇒ y_1 y_2 =h^2 +k^2 −R^2 So x_1 x_2 =y_1 y_2 ((25)/6)=((50)/(b_1 b_2 )) ⇒ b_1 b_2 =12 .

$${Let}\:{eq}.\:{of}\:{the}\:{circle}\:{be} \\ $$$$\:\:\left({x}−{h}\right)^{\mathrm{2}} +\left({y}−{k}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$${for}\:{intercepts}\:{on}\:{x}-{axis}\:{y}=\mathrm{0} \\ $$$$\Rightarrow\:\:\left({x}−{h}\right)^{\mathrm{2}} +{k}^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\:\:\Rightarrow\:{x}_{\mathrm{1}} {x}_{\mathrm{2}} ={h}^{\mathrm{2}} +{k}^{\mathrm{2}} −{R}^{\mathrm{2}} \\ $$$${for}\:{intercepts}\:{on}\:{y}-{axis}\:{x}=\mathrm{0} \\ $$$$\Rightarrow\:\:\left({y}−{k}\right)^{\mathrm{2}} +{h}^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\:\:\Rightarrow\:{y}_{\mathrm{1}} {y}_{\mathrm{2}} ={h}^{\mathrm{2}} +{k}^{\mathrm{2}} −{R}^{\mathrm{2}} \\ $$$$\boldsymbol{{So}}\:\boldsymbol{{x}}_{\mathrm{1}} \boldsymbol{{x}}_{\mathrm{2}} =\boldsymbol{{y}}_{\mathrm{1}} \boldsymbol{{y}}_{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\frac{\mathrm{25}}{\mathrm{6}}=\frac{\mathrm{50}}{\boldsymbol{{b}}_{\mathrm{1}} \boldsymbol{{b}}_{\mathrm{2}} }\:\:\Rightarrow\:\:\:\:\boldsymbol{{b}}_{\mathrm{1}} \boldsymbol{{b}}_{\mathrm{2}} =\mathrm{12}\:. \\ $$$$ \\ $$

Commented by Tinkutara last updated on 11/Jun/18

Thank you Sir.

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