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## The circle passing through(1,-2) and touching the axis of x at (3,0) also passes through ? a)(5,-2) b)(-2,5) c)(-5,2) d)(2,-5)

Ratnaker Mehta
Featured 3 months ago

$\text{(a) } \left(5 , - 2\right)$ is the Right Choice.

#### Explanation:

Let, $r > 0$ is the radius of the circle in Question.

Because the circle touches the X-Axis at $\left(3 , 0\right)$, its

centre C must be, $C = C \left(3 , \pm r\right)$.

Given that the point $P = P \left(1 , - 2\right)$ lies on the circle, we have,

$P {C}^{2} = {r}^{2}$.

#:. (1-3)^2+(+-r+2)^2=r^2...[because," Distance Formula]"#.

$\therefore 4 + \left({r}^{2} \pm 4 r + 4\right) - {r}^{2} = 0$.

$\therefore \pm 4 r = - 8 \Rightarrow r = \pm 2 , \text{ but, as } r > 0 , r = + 2$.

This means that, the centre is $C = C \left(3 , \pm 2\right) , \mathmr{and} , r = 2$.

Hence, the eqns. of the circles are,

${\left(x - 3\right)}^{2} + {\left(y \pm 2\right)}^{2} = {2}^{2} ,$

$\because P \left(1 , - 2\right) \cancel{\in} {\left(x - 3\right)}^{2} + {\left(y - 2\right)}^{2} = {2}^{2}$

$\therefore , \text{the circle under reference is } {\left(x - 3\right)}^{2} + {\left(y + 2\right)}^{2} = {2}^{2}$.

Of all the given points, only $\left(5 , - 2\right)$ satisfies the above eqn.

Hence, $\text{(a) } \left(5 , - 2\right)$ is the Right Choice.

## Find the area of a polygon with the given vertices? (2,5) (7,1) (3,-4) (-2,3) Please show workprocess

sankarankalyanam
Featured 2 months ago

#color(brown)(Area " " ABCD = 20.48 + 19 = 39.48 " sq. units"#

#### Explanation:

$A \left(2 , 5\right) , B \left(7 , 1\right) , C \left(3 , - 4\right) , D \left(- 2 , 3\right)$

Using distance formula,

$\overline{A B} = \sqrt{{\left(7 - 2\right)}^{2} + {\left(1 - 5\right)}^{2}} \approx 6.4$

$\overline{B C} = \sqrt{{\left(3 - 7\right)}^{2} + {\left(- 4 - 1\right)}^{2}} \approx 6.4$

$\overline{C D} = \sqrt{{\left(- 2 - 3\right)}^{2} + {\left(3 + 4\right)}^{2}} \approx 8.6$

$\overline{A D} = \sqrt{{\left(- 2 - 2\right)}^{2} + {\left(3 - 5\right)}^{2}} \approx 4.47$

$\overline{A C} = \sqrt{{\left(3 - 2\right)}^{2} + {\left(- 4 - 5\right)}^{2}} \approx 9.06$

$\text{Area of " Delta " ABC " = sqrt(s * (s - a) * (s - b) * s - c)), " where } s = \frac{a + b + c}{2}$

$s = \frac{6.4 + 6.4 + 9.06}{2} = 10.93$

${\Delta}_{A B C} = \sqrt{10.93 \cdot \left(10.93 - 6.4\right) \cdot \left(10.93 - 6.4\right) \cdot \left(10.93 - 9.06\right)} \approx 20.48$

$\text{Area of " Delta " ADC " = sqrt(s * (s - a) * (s - d) * s - c)), " where } s = \frac{a + d + c}{2}$

$s = \frac{4.47 + 8.6 + 9.06}{2} = 11.065$

${\Delta}_{A D C} = \sqrt{11.065 \cdot \left(11.065 - 4.47\right) \cdot \left(11.065 - 8.6\right) \cdot \left(11.065 - 9.06\right)} \approx 19$

$\text{Area of quadrilateral " ABCD = " Area of " Delta_(ABC) + " Area of } {\Delta}_{A D C}$

#color(brown)(Area " " ABCD = 20.48 + 19 = 39.48 " sq. units"#

## Please solve q 19?

CW
Featured 2 months ago

$D M = \frac{3}{2}$ units

#### Explanation:

Given $\angle D A B = {60}^{\circ} , \angle A D C = {120}^{\circ}$,
$\implies D C$ // $A B$
Let $E$ be the midpoint of $A D$,
$\implies A E = E D = \frac{\sqrt{3}}{2}$
given $M$ is the midpoint of $B C , \implies E M$ // $D C$ // $A B$
$\implies \angle D E M = {60}^{\circ}$,
and $E M = \frac{A B + D C}{2} = \frac{2 \cdot A D}{2} = \frac{2 \sqrt{3}}{2} = \sqrt{3}$
In $\Delta E D M$, as $E M = 2 E D , \mathmr{and} \angle D E M = {60}^{\circ}$,
$\implies \angle E M D = {30}^{\circ} , \implies \angle E D M = {90}^{\circ}$,
$\implies \Delta E D M$ is right angled at $D$,
$\implies D M = E M \sin 60 = \sqrt{3} \cdot \frac{\sqrt{3}}{2} = \frac{3}{2}$ units

Or if you are not sure if $\Delta E D M$ is a right triangle, you can also use the law of cosines to find the length $D M$,
$D {M}^{2} = E {D}^{2} + E {M}^{2} - 2 \cdot E D \cdot E M \cdot \cos 60$,
$\implies D {M}^{2} = {\left(\frac{\sqrt{3}}{2}\right)}^{2} + {\sqrt{3}}^{2} - 2 \cdot \frac{\sqrt{3}}{2} \cdot \sqrt{3} \cdot \frac{1}{2}$
$\implies D {M}^{2} = \frac{3}{4} + 3 - \frac{3}{2} = \frac{9}{4}$
$\implies D M = \sqrt{\frac{9}{4}} = \frac{3}{2}$ units

## What would be the point where the line given in the question below intersects the XY plane?

Ratnaker Mehta
Featured 1 month ago

(d) None of the Above.

#### Explanation:

Clearly, the eqn. of the given line $L$ is,

$L : \left(x , y , z\right) = \left(- 1 , 3 , 3\right) + k \left(1 , 2 , 3\right) , k \in \mathbb{R}$.

In other words, any point on $L$ is,

$\left(- 1 + k , 3 + 2 k , 3 + 3 k\right) , k \in \mathbb{R}$.

Also, the $X Y \text{-plane} = \left\{\left(x , y , 0\right) | x , y \in \mathbb{R}\right\}$.

$\therefore \text{ A point "P in L nn XY"-plane" rArr P in L & P in XY"-plane}$.

$\Rightarrow 3 + 3 k = 0 , \mathmr{and} , k = - 1$.

$\therefore P = P \left(- 1 + k , 3 + 2 k , 3 + 3 k\right)$

$= P \left(- 1 - 1 , 3 - 2 , 3 - 3\right) = P \left(- 2 , 1 , 0\right)$.

Hence, the right choice is (d) None of the Above.

## #2^N# unit spheres are conjoined such that each passes through the center of the opposite sphere. Without using integration, how do you find the common volume?and its limit, as # N to oo#?

Featured 1 month ago

Continuation.

#### Explanation:

The common-volume surface (CVS) is segmented by ${2}^{N}$

equal segments, each reaching the common chord ( axis of

symmetry ) through planar sides, like segments of a peeled orange.

To get a typical segment, rotate the LHS circle in the planar-section

graph ( in my 1st part answer ) about the chord, through $\frac{\pi}{2} ^ \left(N - 1\right)$

rad. The RHS part of the circle generates a segment for the inner

surface, and correspondingly, the larger LHS part forms the

opposite segment, for the outer surface.

Let us study the limit, as $N \to \infty$. The CVS $\to$ ( Rugby ball

shaped ) prolate spheroid of semi-axes

a = b = 1/2 and c = $\frac{\sqrt{3}}{2}$. Its volume is

$\frac{4}{3} \pi \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) \left(\frac{\sqrt{3}}{2}\right) = \pi \frac{\sqrt{3}}{6}$ cu.

The outer surface $\to$ an Earth-like oblate spheroid of semi-

axes a = b = 1.5 and c = 1, with conical dimples at the poles that

are $\left(1 - \frac{\sqrt{3}}{2}\right)$ unit deep.

The volume enclosed, in the limit, is nearly

$\frac{4}{3} \pi \left(\frac{3}{2}\right) \left(\frac{3}{2}\right) \left(1\right) - 2 \left(\frac{1}{3} \pi \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) \left(1 - \frac{\sqrt{3}}{2}\right)\right)$ cu

$= \pi \left(\frac{17}{6} + \frac{\sqrt{3}}{12}\right)$ cu.

Now, the elusive volume of the common-to ${2}^{N}$ conjoined

spheres is expressed as a double integral

V = ${2}^{N}$ ( volume of a typical segment)

$= {2}^{N + 2} \int \int \sqrt{\frac{3}{4} - \rho \cos \theta - {\rho}^{2}} \rho d \theta d \rho$,

with limits

$\rho$ from 0 to $\frac{1}{2} \left(\sqrt{{\cos}^{2} \theta + 3} - \cos \theta\right)$ and

$\theta$ from 0 to $\frac{\pi}{2} ^ N$

I had used cylindrical polar coordinates $\left(\rho , \theta , z\right)$ ,

referred to the common chord as z-axis and its center as origin.

Choosing the center of the LHS sphere as origin, this becomes

$\frac{4}{3} {2}^{N} \int {\left(1 - {\cos}^{2} \alpha {\sec}^{2} \theta\right)}^{\frac{3}{2}} d \theta$, with

$\theta$ from $0 \to \alpha = \frac{\pi}{2} ^ N - {\sin}^{- 1} \left(\frac{1}{2} \sin \left(\frac{\pi}{2} ^ N\right)\right)$.

This is indeed a Gordian knot. So, I look for another method that

leads to a closed form solution. The planar section z = 0 for 8 ( N =

3 ) spheres appears below. The graph reveals most of the aspects

in the description.
graph{((x-0.5)^2+y^2-1)((x+0.5)^2+y^2-1)((y-0.5)^2+x^2-1)((y+0.5)^2+x^2-1)((x-0.3536)^2+(y-0.3536)^2-1)((x+0.3536)^2+(y-0.3536)^2-1)((y+0.3536)^2+(x+0.3536)^2-1)((y+0.3536)^2+(x-0.3536)^2-1)=0[-3 3 -1.5 1.5]}
Now, the common space has just disappeared at z = $\pm \frac{\sqrt{3}}{2}$. This height is $1 \pm \frac{\sqrt{3}}{2}$ above the Table.
graph{((x-0.5)^2+y^2-.25)((x+0.5)^2+y^2-.25)((y-0.5)^2+x^2-.25)((y+0.5)^2+x^2-.25)((x-0.3536)^2+(y-0.3536)^2-.25)((x+0.3536)^2+(y-0.3536)^2-.25)((y+0.3536)^2+(x+0.3536)^2-.25)((y+0.3536)^2+(x-0.3536)^2-.25)=0[-3 3 -1.5 1.5]}
The graphs are on uniform scale.

## What is the orthocenter of a triangle with corners at #(2 ,3 )#, #(6 ,1 )#, and (6 ,3 )#?

maganbhai P.
Featured 1 month ago

Hence, the orthocentre of $\triangle A B C$ is $C \left(6 , 3\right)$

#### Explanation:

Let , $\triangle A B C$ ,be the triangle with corners at

$A \left(2 , 3\right) , B \left(6 , 1\right) \mathmr{and} C \left(6 , 3\right)$ .

We take, $A B = c , B C = a \mathmr{and} C A = b$

So,

${c}^{2} = {\left(2 - 6\right)}^{2} + {\left(3 - 1\right)}^{2} = 16 + 4 = 20$

${a}^{2} = {\left(6 - 6\right)}^{2} + {\left(1 - 3\right)}^{2} = 0 + 4 = 4$

${b}^{2} = {\left(2 - 6\right)}^{2} + {\left(3 - 3\right)}^{2} = 16 + 0 = 16$

It is clear that, ${a}^{2} + {b}^{2} = 4 + 16 = 20 = {c}^{2}$

# i.e. color(red)(c^2=a^2+b^2=>mangleC=pi/2#

Hence, $\overline{A B}$ is the hypotenuse.

$\therefore \triangle A B C$ is the right angled triangle.

$\therefore$The orthocenter coindes with $C$

Hence, the orthocentre of $\triangle A B C$ is $C \left(6 , 3\right)$

Please see the graph:

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