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Equation of Family of Circles Touching a Line and Passing through a Given Point on the Line

For an isosce...

Question

For an isosceles triangle $A B C$ inscribed in a circle of given radius $r$ units, if $h$ is the length of altitude from vertex $A$ to $B C,$ then the value of $lim h \to 0 \frac{Δ}{P^{3}}$ is ( where $Δ, P$ represent area and perimeter of triangle $A B C$ respectively)

\frac{1}{r}

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\frac{1}{64 r}

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\frac{1}{128 r}

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\frac{1}{2 r}

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Solution

The correct option is C $\frac{1}{128 r}$

We know that for an isosceles triangle altidue drawn from vertex and perpendicular bisector of base coincides.

Let, $r =$ radius of circumcircle
$\Rightarrow O A = O B = O C = r$
$B D = \sqrt{B O^{2} - O D^{2}} = \sqrt{r^{2} - {(h - r)}^{2}}$
$= \sqrt{2 r h - h^{2}} \Rightarrow B C = 2 \sqrt{2 r h - h^{2}}$
$A B = \sqrt{B D^{2} + A D^{2}} = \sqrt{2 r h - h^{2} + h^{2}} = \sqrt{2 r h}$
$∴$ Area of triangle $A B C = \frac{1}{2} \times B C \times A D = h \sqrt{2 r h - h^{2}}$
and, Perimeter of triangle $A B C = 2 A B + B C = 2 \sqrt{2 r h} + 2 \sqrt{2 r h - h^{2}}$
Now,

$lim h \to 0 \frac{Δ}{p^{3}} = \frac{h \sqrt{2 r h - h^{2}}}{8 {(\sqrt{2 r h - h^{2}} + \sqrt{2 h r})}^{3}}$
$= lim h \to 0 \frac{h^{3 / 2} \sqrt{2 r - h}}{8 h^{3 / 2} {(\sqrt{2 r - h} + \sqrt{2 r})}^{3}}$
$= \frac{\sqrt{2 r}}{8 {(\sqrt{2 r} + \sqrt{2 r})}^{3}} = \frac{\sqrt{2 r}}{8 \cdot 8 \cdot 2 r \sqrt{2 r}} = \frac{1}{128 r}$

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Equation of Family of Circles Touching a Line and Passing through a Given Point on the Line

Standard XII Mathematics

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For an isosceles triangle ABC inscribed in a circle of given radius r units, if h is the length of altitude from vertex A to BC, then the value of limh→0ΔP3 is ( where Δ,P represent area and perimeter of triangle ABC respectively)

For an isosceles triangle $A B C$ inscribed in a circle of given radius $r$ units, if $h$ is the length of altitude from vertex $A$ to $B C,$ then the value of $lim h \to 0 \frac{Δ}{P^{3}}$ is ( where $Δ, P$ represent area and perimeter of triangle $A B C$ respectively)