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Questions  >>  CBSE XI  >>  Math  >>  Conic Sections
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Q)

State whether the following statement is True or False:The shortest distance from the point $(2,-7)$ to the circle $x^2+y^2-14x-10y-151=0$ is equal to $5$.

[Hint: The shortest distance is equal to the difference of the radius and the distance between the centre and the given point.]

$\begin{array}{1 1}\text{True}\\\text{False}\end{array} $

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A)
Toolbox:
  • The shortest distance is equal to the distance is equal to the difference of the radius and the distance between the centre and the given point.
  • Radius of a circle is $r=\sqrt{g^2+f^2-c}$ where $'g'$ and $'f'$ are the coordinates centre of the circle and $'c'$ is the constant.
  • Distance between two points $A(x_1,y_1)$ and $B(x_2,y_2)$ is given by $AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
Equation of the given circle is $x^2+y^2-14x-10y-151=0$
The given point is $P(2,-7)$
The shortest distance =5
The coordinates of the centre is $(7,5)$
$\therefore$ Radius of the circle is $\sqrt{49+25+151}$
$\Rightarrow \sqrt{225}=15$
Distance between the centre $'O'$ and the point $P(2,-7)$ is
$OP=\sqrt{(7-2)^2+(5+7)^2}$
$\;\;\;\;\;=\sqrt{25+144}$
$\;\;\;\;\;=\sqrt{169}$
$\;\;\;\;\;=13$
Hence the shortest distance $=15-13=2$
Hence the statement is False.
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