Question Number 147009 by gsk2684 last updated on 17/Jul/21
$${a}\:{parabola}\:{y}={x}^{\mathrm{2}} −\mathrm{15}{x}+\mathrm{36}\:{cuts}\:{the}\: \\ $$$${x}\:{axis}\:{at}\:{P}\:\:{and}\:{Q}.\:{a}\:{circle}\:{is}\:{drawn} \\ $$$${through}\:{P}\:{and}\:{Q}\:{so}\:{that}\:{the}\:{origin} \\ $$$${is}\:{outside}\:{it}.\:{then}\:{find}\:{the}\:{length}\: \\ $$$${of}\:{tangent}\:{to}\:{the}\:{circle}\:{from}\:\left(\mathrm{0},\mathrm{0}\right)? \\ $$
Answered by mr W last updated on 17/Jul/21
$${y}={x}^{\mathrm{2}} −\mathrm{15}{x}+\mathrm{36}=\left({x}−\mathrm{3}\right)\left({x}−\mathrm{12}\right. \\ $$$$\Rightarrow{P}\left(\mathrm{3},\mathrm{0}\right) \\ $$$$\Rightarrow{Q}\left(\mathrm{12},\mathrm{0}\right) \\ $$$$\Rightarrow{M}\left(\frac{\mathrm{15}}{\mathrm{2}},\mathrm{0}\right) \\ $$$${let}\:{C}\left({h},\mathrm{0}\right) \\ $$$${OC}^{\mathrm{2}} =\left(\frac{\mathrm{15}}{\mathrm{2}}\right)^{\mathrm{2}} +{h}^{\mathrm{2}} \\ $$$${AC}^{\mathrm{2}} ={CP}^{\mathrm{2}} =\left(\frac{\mathrm{12}−\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} +{h}^{\mathrm{2}} =\left(\frac{\mathrm{9}}{\mathrm{2}}\right)^{\mathrm{2}} +{h}^{\mathrm{2}} \\ $$$${OA}^{\mathrm{2}} ={OC}^{\mathrm{2}} −{AC}^{\mathrm{2}} =\left(\frac{\mathrm{15}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{9}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{OA}=\frac{\sqrt{\mathrm{15}^{\mathrm{2}} −\mathrm{9}^{\mathrm{2}} }}{\mathrm{2}}=\frac{\mathrm{12}}{\mathrm{2}}=\mathrm{6} \\ $$$${that}\:{means}\:{the}\:{length}\:{of}\:{tangent} \\ $$$${is}\:{always}\:\mathrm{6}. \\ $$
Commented by mr W last updated on 17/Jul/21
Commented by mr W last updated on 17/Jul/21
$${method}\:\mathrm{2}: \\ $$$${generally} \\ $$$${OA}^{\mathrm{2}} ={OP}×{OQ}=\mathrm{3}×\mathrm{12}=\mathrm{36} \\ $$$$\Rightarrow{OA}=\mathrm{6} \\ $$$${see}\:{Q}\mathrm{137946} \\ $$
Commented by gsk2684 last updated on 18/Jul/21
$${thank}\:{you} \\ $$