
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
233.4k+ 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.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding the Electric Field of a Uniformly Charged Ring

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions For Class 11 Maths Chapter 12 Limits and Derivatives (2025-26)

NCERT Solutions For Class 11 Maths Chapter 10 Conic Sections (2025-26)

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Derivation of Equation of Trajectory Explained for Students

Understanding Electromagnetic Waves and Their Importance

