
If the lines $3 x-4 y+4=0$ and $6 x-8 y-7=0$ are tangents to a circle, then find the radius of the circle.
A) $\dfrac{3}{4}$
B) $\dfrac{4}{3}$
C) $\dfrac{1}{4}$
D) $\dfrac{7}{4}$
Answer
220.2k+ views
Hint: The radius of a circle is the length of the straight line that connects the centre to any point on its circumference. Because a circle's circumference can contain an endless number of points, a circle can have more than one radius. This indicates that a circle has an endless number of radii and that each radius is equally spaced from the circle's centre. When the radius's length varies, the circle's size also changes.
Complete Step by step solution:
Equation of the given tangents are-
$t_{1}: 3 x-4 y+4=0$
$t_{2}: 6 x-8 y-7=0 \Rightarrow 3 x-4 y-\dfrac{7}{2}=0$
Here,
$a=3, b=-4, c_{1}=4, c_{2}=-\dfrac{7}{2}$
Since the slopes of the supplied tangents are equal in this instance$\dfrac{3}{4}$, they are parallel.
we know about the distance between two parallel lines,
$d=\dfrac{\left|c_{2}-c_{1}\right|}{\sqrt{a^{2}+b^{2}}}$
$\therefore$ distance between ${{\text{t}}_{1}}\text{ }$and ${{\text{t}}_{2}}$$=\dfrac{\left| 4-\left( -\dfrac{7}{2} \right) \right|}{\sqrt{{{3}^{2}}+{{(-4)}^{2}}}}$
$=\dfrac{\left( \dfrac{15}{2} \right)}{\sqrt{9+16}}$
$=\dfrac{15}{2\times 5}=\dfrac{3}{2}$
As we know a circle's diameter is equal to the distance between its two parallel tangents.
$\therefore$ diameter of given circle $=\dfrac{3}{2}$
$\therefore$ Radius of the given circle $=\dfrac{\text { diameter }}{2}=\dfrac{\left(\dfrac{3}{2}\right)}{2}=\dfrac{3}{4}$
As a result, is the specified circle's radius is $\dfrac{3}{4}$.
Hence, the correct answer is option(A).
Note:
A tangent in geometry is a line that originates at an outside point and travels through a curve's centre. When you ride a bicycle, every point on the wheel's circumference establishes a tangent with the road, providing a practical illustration of a tangent.A straight line that only touches or intersects a circle once is said to be its tangent. A line that never enters the interior of the circle is known as a tangent.
Complete Step by step solution:
Equation of the given tangents are-
$t_{1}: 3 x-4 y+4=0$
$t_{2}: 6 x-8 y-7=0 \Rightarrow 3 x-4 y-\dfrac{7}{2}=0$
Here,
$a=3, b=-4, c_{1}=4, c_{2}=-\dfrac{7}{2}$
Since the slopes of the supplied tangents are equal in this instance$\dfrac{3}{4}$, they are parallel.
we know about the distance between two parallel lines,
$d=\dfrac{\left|c_{2}-c_{1}\right|}{\sqrt{a^{2}+b^{2}}}$
$\therefore$ distance between ${{\text{t}}_{1}}\text{ }$and ${{\text{t}}_{2}}$$=\dfrac{\left| 4-\left( -\dfrac{7}{2} \right) \right|}{\sqrt{{{3}^{2}}+{{(-4)}^{2}}}}$
$=\dfrac{\left( \dfrac{15}{2} \right)}{\sqrt{9+16}}$
$=\dfrac{15}{2\times 5}=\dfrac{3}{2}$
As we know a circle's diameter is equal to the distance between its two parallel tangents.
$\therefore$ diameter of given circle $=\dfrac{3}{2}$
$\therefore$ Radius of the given circle $=\dfrac{\text { diameter }}{2}=\dfrac{\left(\dfrac{3}{2}\right)}{2}=\dfrac{3}{4}$
As a result, is the specified circle's radius is $\dfrac{3}{4}$.
Hence, the correct answer is option(A).
Note:
A tangent in geometry is a line that originates at an outside point and travels through a curve's centre. When you ride a bicycle, every point on the wheel's circumference establishes a tangent with the road, providing a practical illustration of a tangent.A straight line that only touches or intersects a circle once is said to be its tangent. A line that never enters the interior of the circle is known as a tangent.
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