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Multiple choice maths fundamental concepts - geometry terms related to polygons curves open and closed figures

The straight line AB is divided at C so that $\bar{AC} = 3\bar{CB}$. Circles are described on AC and CB as diameters and a common tangent meets AB produced at D. Then $\bar{BD}$ equals.

  1. the diameter of the smaller circle

  2. the radius of the smaller circle

  3. the radius of the larger circle
  4. $\bar{CB} \sqrt{3}$

  5. the difference of the two radii
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $ x=\overline{BD} $ and let $ r$ be the radius of the small circle. 


Draw the line from the center of each of the circles to the point of contact of the tangent of the circle. 

By similar triangles, 

$ \dfrac{x+r}{r}=\dfrac{x+5r}{3r} \implies x=r$.

$ \overline{BD} $ equals the radius of the smaller circle.

Multiple choice maths fundamental concepts - geometry terms related to polygons curves open and closed figures

All chords of the curve $x^{2}+y^{2}-10x-4y+4=0$  which make a right angle at $(8,2)$ pass through

  1. $(2,5)$
  2. $(-2,-5)$
  3. $(-5,-2)$
  4. $(5,2)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a circle x^2 + y^2 - 10x - 4y + 4 = 0, the center is (5, 2). Chords subtending a right angle at a point P(8, 2) pass through a fixed point. By the property of circles, if a chord subtends a right angle at P, the locus of the intersection of tangents at the chord's endpoints is related to the circle's geometry. The fixed point is the center (5, 2).

Multiple choice maths fundamental concepts - geometry terms related to polygons curves open and closed figures

Let $m$ be the slope of tangent to the curve $e^{2y}=1+x^{2}$ then set of all values  of $m$ is :

  1. $\left[-\dfrac{1}{2}, \dfrac{1}{2}\right]$
  2. $\left[-\infty, -\dfrac{1}{2}\right]\cup \left[\dfrac{1}{2}, 0\right]$
  3. $\left[-\dfrac{1}{2}, \dfrac{1}{2}\right]-\left\{0\right\}$
  4. $[-2,2]-\left\{0\right\}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$m$ be the slope of tangent.

Given equation of curve is

${{e}^{2y}}=1+{{x}^{2}}$


Taking log both side and we get,

$ \log {{e}^{2y}}=\log \left( 1+{{x}^{2}} \right) $

$ 2y=\log \left( 1+{{x}^{2}} \right) $


On differentiating and we get,

$ 2\dfrac{dy}{dx}=\dfrac{1}{1+{{x}^{2}}}\dfrac{d}{dx}\left( 1+{{x}^{2}} \right) $

$ \Rightarrow 2\dfrac{dy}{dx}=\dfrac{1}{1+{{x}^{2}}}\dfrac{d}{dx}\left( 1+{{x}^{2}} \right) $

$ \Rightarrow 2\dfrac{dy}{dx}=\dfrac{1}{1+{{x}^{2}}}\dfrac{d}{dx}2x $

$ \Rightarrow \dfrac{dy}{dx}=\dfrac{x}{1+{{x}^{2}}} $

$m=\dfrac{dy}{dx}=\dfrac{x}{1+{{x}^{2}}}$


On put $x=\left( -1,1 \right)$

So,

$ m=\dfrac{1}{1+1}=\dfrac{1}{2} $

$ m=\dfrac{-1}{1+1}=\dfrac{-1}{2} $

Hence, the value of $m$ is $\left[-\dfrac{1}{2},\dfrac{1}{2} \right]$


Hence, this is the answer.

Multiple choice maths fundamental concepts - geometry terms related to polygons curves open and closed figures

The radius of the locus by the point represented by $z$, when $arg\dfrac {z-1}{z+1} =\dfrac {\pi}{4}$, is

  1. $\sqrt {2}$
  2. $\sqrt {2}\pi$
  3. $\dfrac {\pi}{\sqrt {2}}$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Arg$\left[\cfrac{z-1}{z+1}\right]=\dfrac{\pi}{4}$

$\Rightarrow$Arg$\left[{z-1}\times \cfrac{1}{z+1}\right]=\dfrac{\pi}{4}$

$\Rightarrow$Arg$\left[{z-1}\right] \times$Arg$\left[\cfrac{1}{z+1}\right]=\dfrac{\pi}{4}$

$\Rightarrow$ Arg${\left[z-1\right]}-$Arg${\left[z+1\right]}=\dfrac{\pi}{4}$

Let $z=x+iy$ 

$\therefore$Arg$\left[\left(x-1\right)+iy\right]-$Arg$\left[\left(x+1\right)+iy\right]=\dfrac{\pi}{4}$

$\Rightarrow {\tan}^{-1}\left(\dfrac{y}{x-1}\right)-{\tan}^{-1}\left(\dfrac{y}{x+1}\right)=\dfrac{\pi}{4}$

$\Rightarrow {\tan}^{-1}{\left(\dfrac{\dfrac{y}{x-1}-\dfrac{y}{x+1}}{1+\dfrac{{y}^{2}}{{x}^{2}-1}}\right)}=\dfrac{\pi}{4}$

$\Rightarrow \dfrac{2y}{{x}^{2}-1+{y}^{2}}=1$

$\Rightarrow {x}^{2}-1+{y}^{2}=2y$

$\Rightarrow {x}^{2}-2y+{y}^{2}-1=0$

This represents a circle with center at $\left(0,1\right)$ and radius  $=\sqrt{0+1-\left(-1\right)}=\sqrt{2}$