Tag: tangents and intersecting chords

Questions Related to tangents and intersecting chords

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

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

  1. $\displaystyle \sqrt{\frac{bc}{a}}$

  2. $ac^2$

  3. $\displaystyle \frac{b}{a}$

  4. $\displaystyle \sqrt{\frac{c}{a}}$

Show answer
Correct Option: D
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 $y=ax^2+bx+c$)

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

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

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

  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. $x^2 + y^2 - 4x - 6y - 12 = 0$

Show answer
Correct Option: A

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

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

  1. $\cfrac { 3 }{ \sqrt { 5 } } $

  2. $\cfrac { \sqrt { 5 } }{ 3 } $

  3. $\sqrt { 5 } $

  4. $\cfrac { 1 }{ \sqrt { 2 } } $

Show answer
Correct Option: B
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 } $

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

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

  1. 5

  2. 10

  3. 15

  4. 20

  5. 25

Show answer
Correct Option: C
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 $k=15$

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 = 

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

  1. 12 cm

  2. 5 cm

  3. 6 cm

  4. 4 cm

Show answer
Correct Option: D
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

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

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

  1. $(1, 3)$

  2. $(3, 1)$

  3. $(1, 2)$

  4. $(2, 1)$

Show answer
Correct Option: B
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.