Question

What is the equation of the circle that touches the line g: 3x - 4y + 8 = 0 at point P (0,2) and goes through point A (7,1)?

Answer 1

The general equation of a circle with centre (a,b) and radius r is (x-a)²+(y-b)²=r².

Equation of circle passing through the point (0,2) is (0-a)²+(2-b)²=r² and point (7,1) is (7-a)²+(1-b)²=r².

Solving the above two equation we get,

7a -b=23 ...(1)

The slope of the given equation 3x-4y+8=0 is 3/4(say m₁). (convert the equation in y=mx+c form i.e., y= (3/4)x+2 ).The slope of the radius is (b-2)/a (say m₂). The line 3x-4y+8=0 touches the circle which implies it is the tangent to the circle. therefore radius is perpendicular to this line.

m₁*m₂=-1

4a+3b=6 ...(2)

solving (1) and (2) we get a=3 and b=-2.

now to find radius (r) we can use the distance of a point (3,-2) from the line 3x-4y+8=0,

r=5

Therefore equation of circle is (x-3)²+(y+2)² =25