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Question Number 79395 by jagoll last updated on 24/Jan/20

given (x,y) is a  point on circle  x^2 +y^2 −6x+4y−23=0.  find minimum and maximum  value of 4x+3y

$$\mathrm{given}\:\left(\mathrm{x},\mathrm{y}\right)\:\mathrm{is}\:\mathrm{a}\:\:\mathrm{point}\:\mathrm{on}\:\mathrm{circle} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{6x}+\mathrm{4y}−\mathrm{23}=\mathrm{0}. \\ $$$$\mathrm{find}\:\mathrm{minimum}\:\mathrm{and}\:\mathrm{maximum} \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{4x}+\mathrm{3y}\: \\ $$

Commented by john santu last updated on 25/Jan/20

⇒(x−3)^2 +(y+2)^2 =23+9+4  (x−3)^2 +(y+2)^2 =36  center at (3,−2), r=6  ⇒let 4x+3y = k ,this is a tangent  to circle .   d=((∣4.3+(−2).3−k∣)/(√(4^2 +3^2 ))), d=r  30 = ∣6−k∣ ⇒6−k=±30  now we get k_(min) =−30  and k_(max) =30

$$\Rightarrow\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left({y}+\mathrm{2}\right)^{\mathrm{2}} =\mathrm{23}+\mathrm{9}+\mathrm{4} \\ $$$$\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left({y}+\mathrm{2}\right)^{\mathrm{2}} =\mathrm{36} \\ $$$${center}\:{at}\:\left(\mathrm{3},−\mathrm{2}\right),\:{r}=\mathrm{6} \\ $$$$\Rightarrow{let}\:\mathrm{4}{x}+\mathrm{3}{y}\:=\:{k}\:,{this}\:{is}\:{a}\:{tangent} \\ $$$${to}\:{circle}\:.\: \\ $$$${d}=\frac{\mid\mathrm{4}.\mathrm{3}+\left(−\mathrm{2}\right).\mathrm{3}−{k}\mid}{\sqrt{\mathrm{4}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }},\:{d}={r} \\ $$$$\mathrm{30}\:=\:\mid\mathrm{6}−{k}\mid\:\Rightarrow\mathrm{6}−{k}=\pm\mathrm{30} \\ $$$${now}\:{we}\:{get}\:{k}_{{min}} =−\mathrm{30} \\ $$$${and}\:{k}_{{max}} =\mathrm{30} \\ $$

Commented by peter frank last updated on 25/Jan/20

thank you

$${thank}\:{you} \\ $$

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