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Question Number 199834 by Calculusboy last updated on 10/Nov/23

Let C be the circle with the center (2,3) and radius 5 a) show that P(5,7) lies on C and find the equation of the tangent at P b) show that the line 3x−4y+31=0 is a tangent to C

$$\boldsymbol{{Let}}\:\boldsymbol{{C}}\:\boldsymbol{{be}}\:\boldsymbol{{the}}\:\boldsymbol{{circle}}\:\boldsymbol{{with}}\:\boldsymbol{{the}}\:\boldsymbol{{center}}\:\left(\mathrm{2},\mathrm{3}\right)\:\boldsymbol{{and}}\:\boldsymbol{{radius}}\:\mathrm{5} \\ $$$$\left.\boldsymbol{{a}}\right)\:\boldsymbol{{show}}\:\boldsymbol{{that}}\:\boldsymbol{{P}}\left(\mathrm{5},\mathrm{7}\right)\:\boldsymbol{{lies}}\:\boldsymbol{{on}}\:\boldsymbol{{C}}\:\boldsymbol{{and}}\:\boldsymbol{{find}}\:\boldsymbol{{the}} \\ $$$$\boldsymbol{{equation}}\:\boldsymbol{{of}}\:\boldsymbol{{the}}\:\boldsymbol{{tangent}}\:\boldsymbol{{at}}\:\boldsymbol{{P}} \\ $$$$\left.\boldsymbol{{b}}\right)\:\boldsymbol{{show}}\:\boldsymbol{{that}}\:\boldsymbol{{the}}\:\boldsymbol{{line}}\:\mathrm{3}\boldsymbol{{x}}−\mathrm{4}\boldsymbol{{y}}+\mathrm{31}=\mathrm{0}\:\boldsymbol{{is}}\:\boldsymbol{{a}}\:\boldsymbol{{tangent}}\:\boldsymbol{{to}}\:\boldsymbol{{C}} \\ $$

Commented by Calculusboy last updated on 10/Nov/23

okay sir

$$\boldsymbol{{okay}}\:\boldsymbol{{sir}} \\ $$

Answered by cortano12 last updated on 10/Nov/23

(a) (5−2)^2 +(7−3)^2 = 9+16=25=5^2 =R^2 so that P(5,7) lies on circle (b) equation of tangent at P(5,7) is (5−2)(x−2)+(7−3)(y−3)=25 3x−6+4y−12=25 3x+4y= 43 (c) the line 3x−4y+31=0 touching the circle so ∣((3.2−4.3+31)/5)∣=((25)/5)=5=R

$$\left(\mathrm{a}\right)\:\left(\mathrm{5}−\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{7}−\mathrm{3}\right)^{\mathrm{2}} =\:\mathrm{9}+\mathrm{16}=\mathrm{25}=\mathrm{5}^{\mathrm{2}} =\mathrm{R}^{\mathrm{2}} \\ $$$$\:\mathrm{so}\:\mathrm{that}\:\mathrm{P}\left(\mathrm{5},\mathrm{7}\right)\:\mathrm{lies}\:\mathrm{on}\:\mathrm{circle} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{equation}\:\mathrm{of}\:\mathrm{tangent}\:\mathrm{at}\:\mathrm{P}\left(\mathrm{5},\mathrm{7}\right) \\ $$$$\:\mathrm{is}\:\left(\mathrm{5}−\mathrm{2}\right)\left(\mathrm{x}−\mathrm{2}\right)+\left(\mathrm{7}−\mathrm{3}\right)\left(\mathrm{y}−\mathrm{3}\right)=\mathrm{25} \\ $$$$\:\mathrm{3x}−\mathrm{6}+\mathrm{4y}−\mathrm{12}=\mathrm{25} \\ $$$$\:\mathrm{3x}+\mathrm{4y}=\:\mathrm{43} \\ $$$$\left(\mathrm{c}\right)\:\mathrm{the}\:\mathrm{line}\:\mathrm{3x}−\mathrm{4y}+\mathrm{31}=\mathrm{0} \\ $$$$\:\mathrm{touching}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{so}\: \\ $$$$\:\:\mid\frac{\mathrm{3}.\mathrm{2}−\mathrm{4}.\mathrm{3}+\mathrm{31}}{\mathrm{5}}\mid=\frac{\mathrm{25}}{\mathrm{5}}=\mathrm{5}=\mathrm{R}\: \\ $$

Commented by Calculusboy last updated on 10/Nov/23

thanks sir

$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$

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