Tag: construction of tangents

The circle equation is (x-1)^2 + (y+2)^2 = 93 + 1 + 4 = 98. The radius is sqrt(98) = 7*sqrt(2). For a square with sides parallel to axes, the distance from center (1, -2) to vertices is the radius.

$\Rightarrow$   Radius of a circle is $r$.

When constructing the circles circumscribing and inscribing a regular hexagon with radius $3$ m, then inscribing hexagon length of each side is $3$ m.

Area of circle $=100\pi $
$\pi r^{2}=100\pi $
$r^{2}=100$
$r=10$
Now, a regular hexagon is made up of 6 equilateral $\bigtriangleup s $ of equal areas. Now, height of equilateral $\bigtriangleup  $ is equal to radius of circle.Therefore, ar. of 1 equilateral $\bigtriangleup=\dfrac {1}{2} $ x base x height
$\Rightarrow \dfrac {\sqrt{3}}{4}a^{2}=\dfrac {1}{2}a*10\Rightarrow a=\dfrac {4*10}{2\sqrt{3}}=\dfrac {20\sqrt{3}}{3} $
Area of hexagon $6\left ( \dfrac {\sqrt{3}}{4}a^{2} \right )=6*\dfrac {\sqrt{3}}{4}\dfrac {20\sqrt{3}}{3}\dfrac {20\sqrt{3}}{3}=200\sqrt{3}$

$AS$ and $AP$ are tangents drawn to the circle at $A$

$\implies AS = AP$

Similarly

$BP = BQ$

$QC = CR$

$RD = DS$

Given

$AD = 23$

$\implies AS + SD = 23$

$AS = 23 – 5 = 18 = AP$

$AB = 29 \implies AP + BP = 29$

$\implies 18 + BP = 29 \implies BP = 11cm$

Now consider rectangle $PBQO$

$PB – BQ , OP = OQ = radius$

$\angle PBQ = 90$    

WKT

$OP \perp BP $ and $OQ \perp BQ$

Since radius is perpendicular to tangent at point of contact

$\implies$ All the angles are 90 degree and adjacent sides are equal

So, It is a square

$\implies r = BP = 11cm$

The tools required for constructing a tangent to a circle is ruler, compass and pencil.