JEE Main 2025 · Mathematics · Straight Lines and Circles · easy

The positive value of k for which the line 3x+4y=k touches the circle x^2+y^2-2x-4y-20=0 is

Numerical answer

The positive value of

k

for which the line

3x+4y=k

touches the circle

x^2+y^2-2x-4y-20=0

Show answer and solution

Answer: 36

The circle is

(x-1)^2+(y-2)^2=25

with centre

(1,2)

and radius

5

. Tangency requires the distance from the centre to the line to equal the radius:

\dfrac{|3+8-k|}{\sqrt{3^2+4^2}}=5

, so

|11-k|=25

. Thus

k=36

k=-14

. The positive value is

36

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