MHT CET · Maths · Circle
If the lines \(3 x-4 y+4=0\) and \(6 x-8 y-7=0\) are tangent to circle, then the radius of the circle is
- A \(\frac{7}{4}\) units
- B \(\frac{3}{4}\) units
- C \(\frac{4}{3}\) units
- D \(\frac{1}{4}\) units
Answer & Solution
Correct Answer
(B) \(\frac{3}{4}\) units
Step-by-step Solution
Detailed explanation
We have, \(3 x-4 y+4=0\)
and \(6 x-8 y-7=0 \Rightarrow 3 x-4 y-\frac{7}{2}=0\)
Lines (1) and (2) are parallel to one another and these lines are tangents to the circle.
\(\therefore\) Distance between the lines is equal to diameter of circle.
\(\therefore \mathrm{D}=\frac{\left|4-\left(\frac{-7}{2}\right)\right|}{\sqrt{(3)^2+(-4)^2}}=\frac{\left(4+\frac{7}{2}\right)}{\sqrt{25}}=\frac{15}{2(5)}=\frac{3}{2} \Rightarrow \text { radius }\)\(=\frac{3}{4} \text { units }\)
and \(6 x-8 y-7=0 \Rightarrow 3 x-4 y-\frac{7}{2}=0\)
Lines (1) and (2) are parallel to one another and these lines are tangents to the circle.
\(\therefore\) Distance between the lines is equal to diameter of circle.
\(\therefore \mathrm{D}=\frac{\left|4-\left(\frac{-7}{2}\right)\right|}{\sqrt{(3)^2+(-4)^2}}=\frac{\left(4+\frac{7}{2}\right)}{\sqrt{25}}=\frac{15}{2(5)}=\frac{3}{2} \Rightarrow \text { radius }\)\(=\frac{3}{4} \text { units }\)
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