Tag: maths

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Questions Related to maths

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

$y=mx+b$ is a tangent to the circle ${x}^{2}+{y}^{2}-6x=16\ if\ \left (3\ m+b\right)^{2}=5\left (1+{m}^{2}\right)$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The condition for a line y = mx + b to be tangent to a circle x^2 + y^2 - 2gx - 2fy + c = 0 is (mg + f + b)^2 = r^2(1 + m^2). For x^2 + y^2 - 6x - 16 = 0, the center is (3, 0) and r^2 = 16 + 9 = 25. Substituting g=3, f=0, r^2=25 gives (3m + b)^2 = 25(1 + m^2), not 5(1+m^2).

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

Let  $ABCD$  be a quadrilateral in which $A B | C D , A B \perp A D \text { and } A B = 3 C D$. The area of quadrilateral  $ABCD$  is  $4.$  The radius of a Circle touching all the sides of quadrilateral is = ?

  1. $\sin \frac { \pi } { 12 }$
  2. $\sin \frac { \pi } { 6 }$
  3. $\sin \frac { \pi } { 4 }$
  4. $\sin \frac { \pi } { 3 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given AB || CD, AB perpendicular to AD, and AB = 3CD, this is a right trapezoid. With area 4, we find the height and side lengths. A circle touches all sides if the sum of opposite sides is equal, which leads to the radius calculation via the geometry of the trapezoid.

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

The tangents drawn from origin to the circle ${ x }^{ 2 }+{ y }^{ 2 }-2ax-2by+{ b }^{ 2 }=0$ are perpendicular to each other, if

  1. $a-b=1$
  2. $a+b=1$
  3. ${ a }^{ 2 }-{ b }^{ 2 }=0$
  4. ${ a }^{ 2 }+{ b }^{ 2 }=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given circle equation ${x}^{2}+{y}^{2}-2ax-2by+{b}^{2}=0$
Center $: (a,b)$ and Radius $= \sqrt { {a}^{2}+{b}^{2}-{b}^{2} } $
Both tangents are drawn from origin and perpendicular to each other. So, two tangent are $x$ and $y$ axis.
Hence, $\sqrt { {a}^{2}+{b}^{2}-{b}^{2} } = a = b$
$\Rightarrow {a}^{2}={b}^{2}$
$\Rightarrow {a}^{2}-{b}^{2} = 0$ 

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

State whether the statement is true/false 

Two tangents $TP$ and $TQ$ are drawn to a circle with center $O$ from an external point $T$, then  $\angle PTQ=\angle OPQ$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In a quadrilateral formed by the center O, the two points of tangency, and the external point T, the angles at the points of tangency are 90 degrees. Thus, angle PTQ + angle POQ = 180 degrees. The statement angle PTQ = angle OPQ is generally false.

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

Opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle. 

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a quadrilateral circumscribing a circle, the triangles formed by the center and the sides have properties such that the central angles subtended by opposite sides sum to 180 degrees.

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

If from a point P, two perpendicular tangents are drawn to the circle ${x^2} + {y^2} - 2x + 2y = 0$, then the coordinates of point P cannot be 

  1. $(3, - 1)$
  2. $(1,1)$
  3. $(\sqrt 3 + 1,0)$
  4. $(2,\sqrt 3 + 1)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The locus of points from which perpendicular tangents are drawn to a circle is the director circle. For x^2 + y^2 - 2x + 2y = 0, the center is (1, -1) and r^2 = 1 + 1 = 2. The director circle is (x-1)^2 + (y+1)^2 = 2(2) = 4. Point (2, sqrt(3)+1) gives (2-1)^2 + (sqrt(3)+1+1)^2 = 1 + (sqrt(3)+2)^2 = 1 + 3 + 4 + 4sqrt(3) = 8 + 4sqrt(3), which is not 4.

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

Let $C _1$ and $C _2$ be two non concentric circles with $C _2$ lying inside $C _1$. A circle C lying inside $C _1$ touches $C _1$ internally and $C _2$ externally. The locus of the centre of the circle C is :

  1. Ellipse

  2. Circle

  3. Parabola

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

Let the radii of C1 and C2 be R and r, and their centers be O1 and O2. If a circle with center (x, y) and radius r' touches C1 internally and C2 externally, the distance to the centers relates to the radii, resulting in the definition of an ellipse.

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

Let $C$ be the circle described $(x+a)^{2}+y^{2}=r^{2}$ where $0<r<a$ Let $m$ be the slope of the line through the origin that is tangent to $C$ at a point in the first quadrant. Then 

  1. $m=\dfrac{r}{\sqrt{a^{2}-r^{2}}}$
  2. $m=\dfrac{\sqrt{a^{2}-r^{2}}}{r}$
  3. $m=\dfrac{r}{a}$
  4. $m=\dfrac{a}{r}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The circle is (x+a)^2 + y^2 = r^2. A line y = mx through the origin is tangent to it. The distance from the center (-a, 0) to the line mx - y = 0 must equal r. Thus, |-ma| / sqrt(m^2 + 1) = r. Squaring gives m^2 a^2 = r^2(m^2 + 1), so m^2(a^2 - r^2) = r^2, leading to m = r / sqrt(a^2 - r^2).

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

Lines are drawn from the point $P(-1,3)$ to the circle $x^{2}+y^{2}-2x+4y-8=0$, which meets the circle at two points A and B. The minimum value of $PA+PB$ is

  1. $4$
  2. $6$
  3. $8$
  4. $16$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$PA+PB \geq 2\sqrt {PA.PB}$   ...{A.M. $\geq $ G.M.}
$PA+PB \geq 2PT$   ...by tangent-secant theorem
$PA+PB \geq 2\sqrt{1+9-(-2)+12-8}=8$

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

A curve is such that the midpoint of the mid-point of the tangent intercepted between the point where the tangent is drawn and the point where the tangent is drawn and the point where the tangent meets y-axis, lies on the line $y=x$. If the curve passes through $(1,0)$, then the curve is

  1. $2y=x^2-x$
  2. $y=x^2-x$
  3. $y=x-x^2$
  4. $y=2(x-x^2)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let $P\left(x,y\right)$ be a point on the curve then equation of the tangent is $Y-y=\dfrac{dy}{dx}\left(X-x\right)$

Given that tangent is drawn and the point where the tangent is drawn and the point where the tangent meets $y$-axis
$\Rightarrow\,x-$coordinate$=0$

$\Rightarrow\,X=0$

$\Rightarrow\, Y-y=\dfrac{dy}{dx}\left(X-x\right)$ becomes

$\Rightarrow\, Y-y=\dfrac{dy}{dx}\left(0-x\right)$ 

$\Rightarrow\, Y=y-x\dfrac{dy}{dx}$

$\therefore\,A=\left(0,y-x\dfrac{dy}{dx}\right)$

Given that  midpoint of the mid-point of the tangent intercepted between the point where the tangent is drawn and the point where the tangent is drawn and the point where the tangent meets $y$-axis, lies on the line $y=x$

$\therefore\,$Midpoint of the line $AP$ lies on the line $y=x$

Midpoint of the line $AP=\left(\dfrac{x+0}{2},\,\dfrac{y+y-x\dfrac{dy}{dx}}{2}\right)$ lies on  the line $y=x$

$\therefore\,x-$coordinate$=y-$coordinate

$\Rightarrow\,\dfrac{x+0}{2}=\dfrac{2y-x\dfrac{dy}{dx}}{2}$

$\Rightarrow\,x=2y-x\dfrac{dy}{dx}$ is a linear differential equation.
$\dfrac{dy}{dx}-\dfrac{2}{x}y=-1$

Integrating factor is $={e}^{\int{p\,dx}}={e}^{\int{\dfrac{-2}{x}\,dx}}={e}^{-\ln{x}}=\dfrac{1}{{x}^{2}}$

Now, $\dfrac{1}{{x}^{2}}\times\dfrac{dy}{dx}-\dfrac{1}{{x}^{2}}\times \dfrac{2}{x}y=\dfrac{-1}{{x}^{2}}$

$\Rightarrow\,\dfrac{1}{{x}^{2}}\dfrac{dy}{dx}-\dfrac{2}{{x}^{3}}y=\dfrac{-1}{{x}^{2}}$ 

$\Rightarrow\,\dfrac{1}{{x}^{2}}dy-\dfrac{2}{{x}^{3}}ydx=\dfrac{-dx}{{x}^{2}}$ 

Integrating both sides,we get

$\Rightarrow\,d\left(\dfrac{y}{{x}^{2}}\right)=d\left(\dfrac{1}{x}\right)$

$\Rightarrow\,d\left(\dfrac{y}{{x}^{2}}\right)=d\left(\dfrac{1}{x}\right)$

$\Rightarrow\,\dfrac{y}{{x}^{2}}=\dfrac{1}{x}+c$ is the required curve.

The curve passes through $\left(1,0\right)$

$\Rightarrow\,0=1+c$

$\Rightarrow\,c=-1$

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

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

$\Rightarrow\,y=x-{x}^{2}$ is the required equation of the curve passing through $\left(1,0\right)$