Question

the line 4x-3y = - 12 is the tangent at point A(-3, 0) and the line 3x+ 4y = 16 is tangent at the point B(4, 1) to a circle. Then equation of the circle is​

Answer 1

Answer:

Step-by-step explanation:

Clearly, centre C of the required circle is the point of intersection of the normals at A and B.

The equation of a line through A (-3,0) and the perpendicular to 4x – 3y = − 12 is - -3

y -0 =

0 ...(i)

Similarly, the equation of a line through B(4,1) and perpendicular to 3x + 4y = 16

is

4 y-1= 0 ...(ii) (x-4) ⇒ 4x − 3y – 13 =

Solving (i) and (ii), we get x = 1, y = -3.

So, the coordinates of Care (1, -3).

...Radius = CA=√16+9 = 5.

Hence, the equation of the required circle is

(x − 1)² + (y + 3)² = 5² or, x² + y² -

2x + 6y - 15 = 0.