Question

Show that Line 3x+4y+7=0 touches the Circle x^2+y^2-4x-6y-12=0

Answer 1

The equation of the given circle can be written as

(x-2)^2+(y-3)^2=25

so the centre of the circle is (2,3) and radius 5 units.

Now calculate the distance of point (2,3) from the line 3x+4y+7=0

which is

(3(2) + 4(3) + 7) \div sqrt{ {3}^{2} + {4}^{2}

equals to 5.

5 is equal to the radius of the circle.

Hence the line is a tangent to the circle.