Question-24379

Question Number 24379 by j.masanja06@gmail.com last updated on 16/Nov/17

Answered by mrW1 last updated on 17/Nov/17

x^2 +y^2 −4x+8y+20=0 (x−2)^2 −4+(y+4)^2 −16+20=0 (x−2)^2 +(y+4)^2 =0 ⇒the circle is in fact a point at (2,−4) please check your question, sir.

$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{8}{y}+\mathrm{20}=\mathrm{0} \\ $$$$\left({x}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{4}+\left({y}+\mathrm{4}\right)^{\mathrm{2}} −\mathrm{16}+\mathrm{20}=\mathrm{0} \\ $$$$\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\left({y}+\mathrm{4}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{the}\:{circle}\:{is}\:{in}\:{fact}\:{a}\:{point}\:{at}\:\left(\mathrm{2},−\mathrm{4}\right) \\ $$$$ \\ $$$${please}\:{check}\:{your}\:{question},\:{sir}. \\ $$

Answered by $@ty@m last updated on 17/Nov/17

The given circle: (x−2)^2 +(y+4)^2 =0 ⇒ it is the point (−2,4) ∴ Equation of tangent y−4=((−3−4)/(5+2))(x+2) y−4=−(x+2) x+y=2

$${The}\:{given}\:{circle}: \\ $$$$\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\left({y}+\mathrm{4}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:{it}\:{is}\:{the}\:{point}\:\left(−\mathrm{2},\mathrm{4}\right) \\ $$$$\therefore\:{Equation}\:{of}\:{tangent} \\ $$$${y}−\mathrm{4}=\frac{−\mathrm{3}−\mathrm{4}}{\mathrm{5}+\mathrm{2}}\left({x}+\mathrm{2}\right) \\ $$$${y}−\mathrm{4}=−\left({x}+\mathrm{2}\right) \\ $$$${x}+{y}=\mathrm{2} \\ $$

Commented by $@ty@m last updated on 17/Nov/17