Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 67071 by TawaTawa last updated on 22/Aug/19

Commented by Prithwish sen last updated on 22/Aug/19

a)The eqn. of C (x−8)^2 +(y−2)^2 =r^2 ∵ L is a tangent ∴ ((k.8−5.2−21)/(√(k^2 +25))) = r ⇒ r^2 = (((8k−31)^2 )/(k^2 +25)) b) ∵ L passes through (18,39) ⇒k.18−5.39−21=0 ⇒k = 12 and r = ((8.12−31)/(√(12^2 +25))) = 5

$$\left.\mathrm{a}\right)\mathrm{The}\:\mathrm{eqn}.\:\mathrm{of}\:\boldsymbol{\mathrm{C}}\:\:\left(\mathrm{x}−\mathrm{8}\right)^{\mathrm{2}} +\left(\mathrm{y}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{r}^{\mathrm{2}} \\ $$$$\because\:\boldsymbol{\mathrm{L}}\:\mathrm{is}\:\mathrm{a}\:\mathrm{tangent} \\ $$$$\therefore\:\frac{\mathrm{k}.\mathrm{8}−\mathrm{5}.\mathrm{2}−\mathrm{21}}{\sqrt{\mathrm{k}^{\mathrm{2}} +\mathrm{25}}}\:=\:\mathrm{r} \\ $$$$\Rightarrow\:\mathrm{r}^{\mathrm{2}} \:=\:\frac{\left(\mathrm{8k}−\mathrm{31}\right)^{\mathrm{2}} }{\mathrm{k}^{\mathrm{2}} +\mathrm{25}} \\ $$$$\left.\mathrm{b}\right)\:\because\:\boldsymbol{\mathrm{L}}\:\mathrm{passes}\:\mathrm{through}\:\left(\mathrm{18},\mathrm{39}\right) \\ $$$$\Rightarrow\mathrm{k}.\mathrm{18}−\mathrm{5}.\mathrm{39}−\mathrm{21}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{k}\:=\:\mathrm{12} \\ $$$$\mathrm{and} \\ $$$$\mathrm{r}\:=\:\frac{\mathrm{8}.\mathrm{12}−\mathrm{31}}{\sqrt{\mathrm{12}^{\mathrm{2}} +\mathrm{25}}}\:=\:\mathrm{5} \\ $$

Commented by TawaTawa last updated on 22/Aug/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by TawaTawa last updated on 22/Aug/19

I appreciate

$$\mathrm{I}\:\mathrm{appreciate} \\ $$

Commented by TawaTawa last updated on 22/Aug/19

Sir, can you please state the formular used ?

$$\mathrm{Sir},\:\mathrm{can}\:\mathrm{you}\:\mathrm{please}\:\mathrm{state}\:\mathrm{the}\:\mathrm{formular}\:\mathrm{used}\:? \\ $$

Commented by TawaTawa last updated on 22/Aug/19

Probably sketch sir

$$\mathrm{Probably}\:\mathrm{sketch}\:\mathrm{sir} \\ $$

Commented by Prithwish sen last updated on 22/Aug/19

a) The equation of a circle with (a,b) as it center and r as its radius is (x−a)^2 +(y−b)^2 =r^2 ∵ L is the tangent . Then the tangent of a circle is always perpendicular to its radius at the same point. So I just equating the perendicular distance from (8,2) to the tangent L with its radius r.

$$\left.\mathrm{a}\right)\:\mathrm{The}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{with}\:\left(\mathrm{a},\mathrm{b}\right)\:\mathrm{as}\:\mathrm{it}\: \\ $$$$\mathrm{center}\:\mathrm{and}\:\mathrm{r}\:\mathrm{as}\:\mathrm{its}\:\mathrm{radius}\:\mathrm{is}\: \\ $$$$\:\:\:\:\left(\mathrm{x}−\mathrm{a}\right)^{\mathrm{2}} +\left(\mathrm{y}−\mathrm{b}\right)^{\mathrm{2}} =\mathrm{r}^{\mathrm{2}} \\ $$$$\because\:\mathrm{L}\:\mathrm{is}\:\mathrm{the}\:\mathrm{tangent}\:.\:\mathrm{Then}\:\mathrm{the}\:\mathrm{tangent}\:\mathrm{of}\:\mathrm{a}\: \\ $$$$\mathrm{circle}\:\mathrm{is}\:\mathrm{always}\:\mathrm{perpendicular}\:\mathrm{to}\:\mathrm{its}\:\mathrm{radius} \\ $$$$\mathrm{at}\:\mathrm{the}\:\mathrm{same}\:\mathrm{point}.\:\mathrm{So}\:\mathrm{I}\:\mathrm{just}\:\mathrm{equating}\:\mathrm{the}\: \\ $$$$\mathrm{perendicular}\:\mathrm{distance}\:\mathrm{from}\:\left(\mathrm{8},\mathrm{2}\right)\:\mathrm{to}\:\mathrm{the}\: \\ $$$$\mathrm{tangent}\:\mathrm{L}\:\mathrm{with}\:\mathrm{its}\:\mathrm{radius}\:\mathrm{r}. \\ $$

Commented by Prithwish sen last updated on 22/Aug/19

Terms of Service

Privacy Policy

Contact: info@tinkutara.com