Mathematics · Quantitative Aptitude

Circle and Arc Properties

115 Questions

Circle and arc properties involve calculating arc lengths, understanding radius relationships, and solving geometric proofs. These geometry concepts are crucial for quantitative aptitude tests. Review these questions to improve your spatial reasoning and accuracy.

Arc length calculationsCircle theoremsRadius and diameterInscribed polygonsCentral angles

Circle and Arc Properties Questions

Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

If $r$ and $R$ are respectively the radii of the inscribed and circumscribed circles of a regular polygon of $n$ sides such that $\dfrac{R}{r}=\sqrt{5}-1$, then $n$ is equal to

  1. $5$
  2. $6$
  3. $10$
  4. $18$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $a$ be the length of side of regular polygon then
$R=\dfrac{a}{2}\csc{\left(\dfrac{\pi}{n}\right)}$ and $r=\dfrac{a}{2}\cot{\left(\dfrac{\pi}{n}\right)}$
$\therefore \dfrac{R}{r}=\dfrac{\csc{\left(\dfrac{\pi}{n}\right)}}{\cot{\left(\dfrac{\pi}{n}\right)}}=\dfrac{\dfrac{1}{\sin\left(\frac{\pi}{n}\right)}}{\dfrac{\cos{\frac{\pi}{n}}}{\sin{\frac{\pi}{n}}}}=\dfrac{1}{\cos{\left(\frac{\pi}{n}\right)}}$
$\therefore \cos{\left(\dfrac{\pi}{n}\right)}=\dfrac{1}{\sqrt{5}-1}$
On rationalising the denominator, we get 
$ \cos{\left(\dfrac{\pi}{n}\right)}=\dfrac{1}{\sqrt{5}-1}=\dfrac{1}{\sqrt{5}-1}\times\dfrac{\sqrt{5}-1}{\sqrt{5}+1}=\dfrac{\sqrt{5}+1}{4}=\cos{\left(\dfrac{\pi}{5}\right)}$
On comparing $\cos{\left(\dfrac{\pi}{n}\right)}=\cos{\left(\dfrac{\pi}{5}\right)}$ we get $n=5$

Multiple choice mathematics and statistics hyperbola auxiliary circle auxiliary circle, director circle director circle and auxiliary circle of a hyperbola

The length of diameter of director circle of hyperbola $\dfrac{x^2}{49}-\dfrac{y^2}{25}=1$, is 

  1. $4$
  2. $6$
  3. $4\sqrt6$
  4. $24$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The Director circle of a hyperbola is defined as the locus of the point of intersection of two perpendicular tangents to the hyperbola. For any standard hyperbola $\dfrac{x^2}{a^2} -\dfrac {y^2}{b^2} = 1$,


The equation of Director circle is given by $x^2 + y^2 = a^2 - b^2$


Here the given hyperbola is $\dfrac{x^2}{49} -\dfrac {y^2}{25} = 1$,

Here $a =7$ and $b =5$

So equation of the director circle will be $x^2 + y^2 = (7)^2 - (5)^2$

$\Rightarrow x^2 + y^2 = 24$

Hence the radius of the director circle is $\sqrt{24} = 2\sqrt{6}$. So the diameter will be $4\sqrt6$.

So correct option is $C$.

Multiple choice mathematics and statistics hyperbola auxiliary circle auxiliary circle, director circle director circle and auxiliary circle of a hyperbola

The radius of the director circle of the hyperbola $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$ is 

  1. $a-b$
  2. $\sqrt{a-b}$
  3. $\sqrt{a^{2}-b^{2}}$
  4. $\sqrt{a^{2}+b^{2}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The locus of the point of intersection of perpendicular tangents to a conic is called its director circle. For a standard hyperbola x^2/a^2 - y^2/b^2 = 1, the equation of the director circle is x^2 + y^2 = a^2 - b^2, provided a^2 > b^2. Therefore, the radius of this director circle is the square root of a^2 - b^2.

Multiple choice mathematics and statistics hyperbola auxiliary circle auxiliary circle, director circle director circle and auxiliary circle of a hyperbola

Auxiliary circle of a hyperbola is defined as:

  1. The auxiliary circle for a hyperbola is a circle with its centre on the polar and contains the two vertices.

  2. The circle whose center concurs with that of the ellipse and whose radius is equal to the ellipse's semimajor axis.

  3. The auxiliary circle for a hyperbola is a circle with its centre on the axis and contains the two vertices.

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The Auxiliary circle of a hyperbola is defined as the circle with the center same as the hyperbola and with transverse axis as it's diameter. The end points of transverse axis are the two vertices of the hyperbola, so the circle also contains the two vertices of the hyperbola.


The equation of Auxiliary circle foe a hyperbola is given by $x^2  + y^2 = a^2$

Hence the correct answer is Option $C$.

Multiple choice mathematics and statistics hyperbola auxiliary circle auxiliary circle, director circle director circle and auxiliary circle of a hyperbola

The radius of director circle of the hyperbola $\dfrac{x^2}{16}-\dfrac{y^2}{9}=1$ is

  1. $6$
  2. $7$
  3. $\sqrt 7$
  4. $8$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The Director circle of a hyperbola is defined as the locus of the point of intersection of two perpendicular tangents to the hyperbola. For any standard hyperbola $\dfrac{x^2}{a^2} -\dfrac {y^2}{b^2} = 1$,


The equation of Director circle is given by $x^2 + y^2 = a^2 - b^2$


Here the given hyperbola is $\dfrac{x^2}{16} -\dfrac {y^2}{9} = 1$,

Here $a =4$ and $b =3$

So equation of the director circle will be $x^2 + y^2 = (4)^2 - (3)^2$

$\rightarrow x^2 + y^2 = 7$

Hence the radius of the director circle is $\sqrt7$. So correct option is $C$.

Multiple choice mathematics and statistics hyperbola auxiliary circle auxiliary circle, director circle director circle and auxiliary circle of a hyperbola

The radius of  director circle of hyperbola is $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$

  1. $a$
  2. $b$
  3. $\sqrt{a^2+b^2}$
  4. $\sqrt{a^2-b^2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The Director circle of a hyperbola is defined as the locus of the point of intersection of two perpendicular tangents to the hyperbola. For any standard hyperbola $\dfrac{x^2}{a^2} -\dfrac {y^2}{b^2} = 1$,


The equation of Director circle is given by $x^2 + y^2 = a^2 - b^2$


Hence the Director circle is a circle whose centre is same as centre of the hyperbola and the radius is $\sqrt{a^2 - b^2}$

So correct option is $D$.

Multiple choice mathematics and statistics hyperbola auxiliary circle auxiliary circle, director circle director circle and auxiliary circle of a hyperbola

The director circle intersects its hyperbola in _______ number of points.

  1. zero

  2. two

  3. three

  4. four
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a standard hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$, the equation of director circle is given by:


$x^2 + y^2 = a^2-b^2$

So we can see the center of the director circle of the standard hyperbola is same as the hyperbola. the radius of director circle is $\sqrt{a^2-b^2}$

As $a^2>0$, $a^2 > a^2 -b^2$

or $a > \sqrt{a^2-b^2}$

So we can see the radius of director circle is less the vertex of the hyperbola. Hence with a center same as the hyperbola, the director circle doesn't cut the hyperbola at any point. or we can say it's just smaller than the Auxiliary circle of the hyperbola which touches the vertices of the hyperbola. 

Hence the correct option is $A$.

Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

If $\dfrac { z+2i }{ z-2i } $ is purely imaginary then $\left| z \right| $ is 

  1. $1$
  2. $2$
  3. $\dfrac { 1 }{ 2 } $
  4. $\dfrac { 1 }{ 4 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\dfrac{z+2i}{z-2i}$

$=\dfrac{z+2i}{z-2i}\times \dfrac{z+2i}{z+2i}$

$=\dfrac{z^2+4z-4}{z^2-4}$

Let $z=x+iy$

$=\dfrac{(x+iy)^2+4(x+iy)-4}{(x+iy)^2-4}$

$=\dfrac{x^2-y^2+2ixy-4+4x+4iy}{(x+iy)^2-4}$

$=\dfrac{(x^2-y^2+4x-4)+i(2xy+4y)}{(x+iy)^2-4}$

z is purely imaginary. So, Re(z)=0

$\dfrac{x^2-y^2+4x-4}{(x+iy)^2-4}=0$

$\Rightarrow x^2-y^2+4x-4=0$

$\therefore x^2-y^2+4x=4$

The above equation represents the hyperbola on x-axis, with $(1,0)$

$\therefore z=1$
Multiple choice maths circle measures area between two concentric circles the area of ring semicircle and ring

In a triangle with sides $a$, $b$, and $c$, a semicircle touching the sides $AC$ and $CB$ is inscribed whose diameter lies on $AB$. Then the radius of the semicircle is

  1. $a/2$
  2. $\triangle/s$
  3. $\dfrac{2\triangle}{a+b}$
  4. $\dfrac{2\ abc}{(s)(a+b)}\cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The area of the triangle can be split into two triangles formed by the center of the semicircle and sides AC and CB, with heights equal to the radius r. Thus, Area = (1/2) a r + (1/2) b r = (r/2)(a + b), which rearranges to r = 2(Area)/(a + b).

Multiple choice maths circle measures area between two concentric circles the area of ring semicircle and ring

Points $P,Q,R$ lie on same line. Three semi circles with the diameters $PQ,QR,PR$ are drawn on same side of line segment $PR$. The centres of the semicircles are $A,B,O$ respectively. A circle with centre $C$ touches all $3$ semi circles then the radius of this circle is $\left(AQ=a,BQ=b\right)$

  1. $\dfrac{ab}{a+b}$
  2. $\dfrac{ab\left(a+b\right)}{a^{2}+b^{2}}$
  3. $\dfrac{ab\left(a+b\right)}{a^{2}+ab+b^{2}}$
  4. $\dfrac{ab\left(a+b\right)}{\left(a-b\right)^{2}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a classic geometry problem involving the Sangaku theorem or properties of circles tangent to a line. The radius of the circle tangent to three semicircles with diameters a, b, and a+b is given by r = (a * b) / (a + b).

Multiple choice maths circle measures area between two concentric circles the area of ring semicircle and ring

A semicircle is drawn with $AB$ as its diameter. From $C$ a point on $AB$ a line perpendicular to $AB$ is drawn meeting the circumference of the semicircle at $D$. Given that $AC = 2\ cm$ and $CD = 6\ cm$ the area of the semicircle is :

  1. $\displaystyle 32\pi $
  2. $\displaystyle 50\pi $
  3. $\displaystyle 40\pi $
  4. $\displaystyle 36\pi $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let O be the centre of the circle. Then, $OA = OB = OD= r$
Now, $OC = r - 2$
and $CD = 6$
Thus, in $\triangle ODC$
$OC^2 + CD^2 = OD^2$
$(r - 2)^2 + 6^2 = r^2$
$r^2 + 4 - 4r + 36 = r^2$
$4r = 40$
$r = 10$ $cm$
Area of semicircle $= \dfrac{\pi r^2}{2} = \dfrac{\pi (10)^2}{2} = 50 \pi$

Multiple choice maths circle measures area between two concentric circles the area of ring semicircle and ring

Jackson measured the button on his shirt. Then he calculated that it has a semicircle of $25.12\ mm$. What is the button's radius? (Use $\pi = 3.14$).

  1. $1\ mm$
  2. $2\ mm$
  3. $3\ mm$
  4. $4\ mm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Area of a semicircle $=\dfrac{1}{2}\pi r^2$
$ 25.12= \dfrac{1}{2}\pi r^2$
$ 50.24= \pi r^2  $

Using $\pi = 3.14$ (given)
$16 = r^2$
$r = 4\ mm$

Multiple choice maths circle measures area between two concentric circles the area of ring semicircle and ring

The area in ( ${cm^2}$) of the largest triangle that can be inscribed in a semicircle of radius r cm is 

  1. ${\cfrac{1}{3}\pi r^2}$
  2. ${2r^2}$
  3. ${r^2}$
  4. ${\cfrac{1}{2}\pi r^2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The largest triangle inscribed in a semicircle has a height = radius of the circle

base = diameter of the circle
$\therefore$ Area of a triangle $=\dfrac{1}{2}\times base\times height$
                                  $=\dfrac{1}{2}\times 2r \times r=r^2$
Hence, option C is correct.

Multiple choice maths circle measures area between two concentric circles the area of ring semicircle and ring

A wire in the shape of an equilateral triangle encloses an area $s$ sq. cm  If the same wire is bent to form circle, the area of the circle will be

  1. $\displaystyle \frac{\pi s^{2}}{9}$
  2. $\displaystyle \frac{3s^{2}}{\pi }$
  3. $\displaystyle \frac{3s}{\pi }$
  4. $\displaystyle \frac{3\sqrt{3}s}{\pi }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Area of equilateral triangle $= s$ sq.cm
$\Rightarrow  \dfrac{\sqrt3}{4} a^2 = s$, [where $a$, the side of equilateral triangle]
$\Rightarrow a= \sqrt{\dfrac{4s}{\sqrt3}}$
Now perimeter of equilateral triangle $ 3\times a =3 \times\sqrt{\dfrac{4s}{\sqrt 3}}$ cm 
Circumference of circle $=$ perimeter of equilateral triangle
$\Rightarrow 2\pi r= 3 \times\sqrt{\dfrac{4s}{\sqrt 3}}$, [where $r$ the radius of circle]
Solve the above expression for $r$, we get 
$r= \dfrac{3}{2\pi} \times \sqrt{\dfrac{4s}{\sqrt 3}}$
Area of circle $=\pi r^2 = \pi \times \left ( \dfrac{3}{2\pi} \times \sqrt{\dfrac{4s}{\sqrt 3}} \right )^2$
After simplification, we get
Area of circle $=\dfrac{3s\sqrt3}{\pi}$ sq.cm