Question and Answers Forum

All Questions      Topic List

Coordinate Geometry Questions

Previous in All Question      Next in All Question      

Previous in Coordinate Geometry      Next in Coordinate Geometry      

Question Number 187142 by a.lgnaoui last updated on 14/Feb/23

 Montrer que:  4r^2 =a^2 +b^2 +c^2 +d^2

$$\:{Montrer}\:{que}: \\ $$$$\mathrm{4}{r}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \\ $$$$ \\ $$

Commented by a.lgnaoui last updated on 14/Feb/23

Commented by mahdipoor last updated on 14/Feb/23

C(O,r) :   x^2 +y^2 =r^2   P=(i,j) { ((line CPD : x=i)),((line APB : y=j)) :}  A=(−(√(r^2 −j^2 )),j)        B=(+(√(r^2 −j^2 )),j)  C=(i,(√(r^2 −i^2 )))              D=(i,−(√(r^2 −i^2 )))  a^2 +b^2 +c^2 +d^2 =∣AP∣^2 +∣AB∣^2 +...=  [(i−(−(√(r^2 −j^2 ))))^2 +(j−j)^2 ]+[(i−(√(r^2 −j^2 )))^2 +(j−j)^2 ]  +...=2i^2 +2(r^2 −j^2 )+2j^2 +2(r^2 −i^2 )=4r^2

$${C}\left({O},{r}\right)\::\:\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${P}=\left({i},{j}\right)\begin{cases}{{line}\:{CPD}\::\:{x}={i}}\\{{line}\:{APB}\::\:{y}={j}}\end{cases} \\ $$$${A}=\left(−\sqrt{{r}^{\mathrm{2}} −{j}^{\mathrm{2}} },{j}\right)\:\:\:\:\:\:\:\:{B}=\left(+\sqrt{{r}^{\mathrm{2}} −{j}^{\mathrm{2}} },{j}\right) \\ $$$${C}=\left({i},\sqrt{{r}^{\mathrm{2}} −{i}^{\mathrm{2}} }\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:{D}=\left({i},−\sqrt{{r}^{\mathrm{2}} −{i}^{\mathrm{2}} }\right) \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} =\mid{AP}\mid^{\mathrm{2}} +\mid{AB}\mid^{\mathrm{2}} +...= \\ $$$$\left[\left({i}−\left(−\sqrt{{r}^{\mathrm{2}} −{j}^{\mathrm{2}} }\right)\right)^{\mathrm{2}} +\left({j}−{j}\right)^{\mathrm{2}} \right]+\left[\left({i}−\sqrt{{r}^{\mathrm{2}} −{j}^{\mathrm{2}} }\right)^{\mathrm{2}} +\left({j}−{j}\right)^{\mathrm{2}} \right] \\ $$$$+...=\mathrm{2}{i}^{\mathrm{2}} +\mathrm{2}\left({r}^{\mathrm{2}} −{j}^{\mathrm{2}} \right)+\mathrm{2}{j}^{\mathrm{2}} +\mathrm{2}\left({r}^{\mathrm{2}} −{i}^{\mathrm{2}} \right)=\mathrm{4}{r}^{\mathrm{2}} \\ $$

Answered by mr W last updated on 14/Feb/23

(b−((a+b)/2))^2 +(((c+d)/2))^2 =r^2   ⇒(a−b)^2 +(c+d)^2 =4r^2    ...(i)  (d−((c+d)/2))^2 +(((a+b)/2))^2 =r^2    ⇒(c−d)^2 +(a+b)^2 =4r^2    ...(ii)  (i)−(ii):  4cd−4ab=0 ⇒cd=ab  (i):  a^2 +b^2 −2ab+c^2 +d^2 +2cd=4r^2   ⇒a^2 +b^2 +c^2 +d^2 =4r^2   ✓

$$\left({b}−\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{c}+{d}}{\mathrm{2}}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\Rightarrow\left({a}−{b}\right)^{\mathrm{2}} +\left({c}+{d}\right)^{\mathrm{2}} =\mathrm{4}{r}^{\mathrm{2}} \:\:\:...\left({i}\right) \\ $$$$\left({d}−\frac{{c}+{d}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \: \\ $$$$\Rightarrow\left({c}−{d}\right)^{\mathrm{2}} +\left({a}+{b}\right)^{\mathrm{2}} =\mathrm{4}{r}^{\mathrm{2}} \:\:\:...\left({ii}\right) \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$$\mathrm{4}{cd}−\mathrm{4}{ab}=\mathrm{0}\:\Rightarrow{cd}={ab} \\ $$$$\left({i}\right): \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}+{c}^{\mathrm{2}} +{d}^{\mathrm{2}} +\mathrm{2}{cd}=\mathrm{4}{r}^{\mathrm{2}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} =\mathrm{4}{r}^{\mathrm{2}} \:\:\checkmark \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com