Question

Show that the line 3x + 4y + 7 = 0 touches the circle x² + y² – 4x – 6y – 12 = 0 and find the point of contact.

Answer 1

Given centre = (2, 3) and r = \(\sqrt{4 + 9 + 12}\) = √25 = 5 

If the length of the perpendicular is equal to radius of the circle than the line touches the circle

∴ The line 3x + 4y + 7 touches the circle 

Any line perpendicular to 3x + 4y + 7 = 0 is of the form 4x – 3y + k = 0 

It passes through (2,3) 

∴ 8 – 9 + k = 0 ⇒ k = 1 

∴ The line is 4x – 3y + 1 = 0 

Point of contact is the intersection of 3x + 4y + 7 = 0 and 4x – 3y + 1 = 0

∴ The point of contact is (-1,-1)