find-the-equation-of-the-circle-which-ends-one-of-the-diameters-of-two-points-p-1-2-3-and-p-2-4-5-

Question Number 69462 by mhmd last updated on 23/Sep/19

$$ \\ $$$${find}\:{the}\:{equation}\:{of}\:{the}\:{circle}\:{which}\:{ends}\:{one}\:{of}\:{the}\:{diameters}\:{of}\:{two}\:{points}\:{p}_{\mathrm{1}} \left(−\mathrm{2},\mathrm{3}\right)\:{and}\:{p}_{\mathrm{2}} \left(\mathrm{4},\mathrm{5}\right) \\ $$

Commented by mathmax by abdo last updated on 23/Sep/19

let m(x,y) m∈ C ⇔mp_1 ^→ .mp_2 ^→ =0 ⇒ (((−2−x)),((3−y)) ). (((4−x)),((5−y)) )=0 ⇒ (−2−x)(4−x)+(3−y)(5−y) =0 ⇒−8+2x−4x+x^2 +15−3y −5y +y^2 =0 ⇒x^2 +y^2 −2x−8y +7 =0 ⇒ C : x^2 +y^2 −2x−8y +7 =0

$${let}\:{m}\left({x},{y}\right)\:\:{m}\in\:{C}\:\Leftrightarrow{m}\overset{\rightarrow} {{p}}_{\mathrm{1}} .{m}\overset{\rightarrow} {{p}}_{\mathrm{2}} =\mathrm{0}\:\Rightarrow\begin{pmatrix}{−\mathrm{2}−{x}}\\{\mathrm{3}−{y}}\end{pmatrix}.\begin{pmatrix}{\mathrm{4}−{x}}\\{\mathrm{5}−{y}}\end{pmatrix}=\mathrm{0}\:\Rightarrow \\ $$$$\left(−\mathrm{2}−{x}\right)\left(\mathrm{4}−{x}\right)+\left(\mathrm{3}−{y}\right)\left(\mathrm{5}−{y}\right)\:=\mathrm{0}\:\Rightarrow−\mathrm{8}+\mathrm{2}{x}−\mathrm{4}{x}+{x}^{\mathrm{2}} \\ $$$$+\mathrm{15}−\mathrm{3}{y}\:−\mathrm{5}{y}\:+{y}^{\mathrm{2}} \:=\mathrm{0}\:\Rightarrow{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{8}{y}\:+\mathrm{7}\:=\mathrm{0}\:\Rightarrow \\ $$$${C}\:\::\:\:\:\:{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{8}{y}\:+\mathrm{7}\:=\mathrm{0} \\ $$

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