Tag: tangents and intersecting chords

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Questions Related to tangents and intersecting chords

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

A parabola $y = ax^2 + bx + c$ crosses the x-axis at $(\alpha, 0)$ $(\beta, 0)$ both to the right of the origin. A circle also passes through these two points. The length of the tangent from the origin to the circle is

  1. $\displaystyle \sqrt{\frac{bc}{a}}$
  2. $ac^2$
  3. $\displaystyle \frac{b}{a}$
  4. $\displaystyle \sqrt{\frac{c}{a}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$OT$ is a tangent and $OAB$ is a secant 


we know that

$OT^2 =OA.OB$

         $=\alpha\beta$

         $=\dfrac{c}{a}$ (Since $\alpha,\beta $ are the roots of $y=ax^2+bx+c$)

$\Rightarrow OT=\sqrt{\dfrac{c}{a}}$

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

From a point $R(5, 8)$ two tangents $RP$ and $RQ$ are drawn to a given cirlce $S = 0$ whose radius is $5$. If circumcentre of the triangle PQR is $(2, 3)$, then the equation of circle $S= 0$ is

  1. $x^2 + y^2 + 2x + 4y - 20 = 0$
  2. $x^2 + y^2 + x + 2y - 10 = 0$
  3. $x^2 + y^2 - x - 2y - 20 = 0$
  4. $x^2 + y^2 - 4x - 6y - 12 = 0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The circumcenter of triangle PQR, where P and Q are points of tangency from R, is the midpoint of the chord of contact and the center of the circle. Using the given circumcenter and radius, the equation can be derived.

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

The radius of the circle touching the straight lines $x-2y-1=0$ and $3x-6y+7=0$ is

  1. $\cfrac { 3 }{ \sqrt { 5 } } $
  2. $\cfrac { \sqrt { 5 } }{ 3 } $
  3. $\sqrt { 5 } $
  4. $\cfrac { 1 }{ \sqrt { 2 } } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Diameter of circle=distance of the point (1,0)
from $3x-6y+7=0$
$\therefore$ $\cfrac { 3(1)-6(0)+7 }{ \sqrt { { \left( 3 \right)  }^{ 2 }+{ \left( -6 \right)  }^{ 2 } }  } =\cfrac { 10 }{ \sqrt { 45 }  } =\cfrac { 2 }{ 3 } \sqrt { 5 } $
Now, radius of circle $=\cfrac { 1 }{ 2 } \left( \cfrac { 2 }{ 3 } \sqrt { 5 }  \right) =\cfrac { \sqrt { 5 }  }{ 3 } $

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

For what positive value(s) of K will the graph of the equation $2x + y = K$ be tangent to the graph of the equation $x^2+ y^2= 45$?

  1. 5

  2. 10

  3. 15

  4. 20

  5. 25
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
  • The radius of circle is $\sqrt{45} = 3\sqrt5$ , center of circle is $(0,0)$
  • For the equation to be tangent to circle , the distance from center of circle to given line must be equal to radius of circle
  • So we get $k/\sqrt5 = 3\sqrt5$ , which gives $k=15$
Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

AB and CD are two chords of a circle which when produced to meet at a point P such that AB = 5 cm, AP = 8 cm and CD = 2 cm then PD = 

  1. 12 cm

  2. 5 cm

  3. 6 cm

  4. 4 cm
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

By intersecting secant theorem,

$PA$$\times$$PB$ = $PD$$\times$$PC$
$8$cm$\times$$3$cm = PD$\times$(PD+CD)
24${ cm }^{ 2 }$ = PD$\times$(PD+2)
${ PD }^{ 2 }$ $+ 2PD - 24 =0$
On Solving the above quadratic equation, we get
${ PD }^{ 2 }$$+6PD-4PD-24=0$
$(PD+6)$$\times$$(PD-4)=0$
$PD=4$cm & $-6$cm
So, $PD= 4$cm is the real solution

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

If the line $\displaystyle ax+by + c =0$ touches the circle $\displaystyle x^2 + y^2 -2x = \frac{3}{5}$ and is normal to the circle $\displaystyle x^2 + y^2 + 2x - 4y + 1 =0$, then $(a,b)$ are

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

$x^2+y^2-2x=\dfrac {3}{5}\Rightarrow (x-1)^2+y^2=\dfrac {8}{5}$

So, Radius, $R=2\sqrt {\dfrac {2}{5}}$ and it's center is at $(1,0)$

ie, Distance, $d$ from the circle to $ax+by+c=0$ is,
$d=\dfrac {a\times 1+b\times 0+c}{\sqrt{a^2+b^2}}=\dfrac {a+c}{\sqrt{a^2+b^2}} =2\sqrt {\dfrac {2}{5}}\longrightarrow (1)$ (Inorder to satisfy the criterion of a tangent)

$x^2+y^2+2x-4y+1=0 \Rightarrow (x+1)^2+(y-2)^2=4$
So, It's center is at $((-1),2)$
As $ax+by+c=0$ is normal to the circle, it should go through the centre of the circle.
ie, $a-2b=c$ and $(y-2)=m(x+1)\longrightarrow (2)$

Substituting $c$ in (1),
$\dfrac {a+(a-2b)}{\sqrt{a^2+b^2}} =2\sqrt {\dfrac {2}{5}}$
$\Rightarrow \dfrac {a-b}{\sqrt {a^2+b^2}}=\sqrt {\dfrac {2}{5}}$

So, we can say $(a-b)=k\sqrt {2}$ and $a^2+b^2=5k^2$ foe some constant $k$.
$a^2+b^2-(a-b)^2=2ab=5k^2-2k^2=3k^2$
$(a-b)^2+4ab=(a+b)^2=6k^2+2k^2=8k^2\Rightarrow (a+b)=2k\sqrt{2}$
$a=\dfrac {1}{2}((a+b)+(a-b))=\dfrac {1}{2}(3k\sqrt{2})$
$b=\dfrac {1}{2}((a+b)-(a-b))=\dfrac {1}{2}(k\sqrt {2})$

Slope of the line, $m=\dfrac {dy}{dx}$
$\dfrac {d}{dx}(ax+by+c)=0\Rightarrow a+b\dfrac {dy}{dx}=0$
ie, $m=\dfrac {(-a)}{b}=(-3)$ (from above equations of $a$ and $b$)

Substituting the slope in (2),
$(y-2)=(-3)(x+1)\Rightarrow 3x+y+1=0$

Compairing with general equation given,
$(a,b)=(3,1)$

Option B is the correct answer.