a. The radius of the circle is 5.
b. Since the gradient of the normal line is the negative reciprocal of the tangent line's gradient, the gradient of the normal line at the point (3, -4) is -4/3.
How to find the radius of the circle
(a) To find the radius of the circle given the equation x² + y² = a² and the point (3, -4) lying on the circle, substitute the coordinates of the point into the equation and solve for a.
Substitute x = 3 and y = -4 into the equation:
3² + (-4)² = a²
9 + 16 = a²
25 = a²
Taking the square root of both sides to solve for a:
a = √25
a = 5
Therefore, the radius of the circle is 5.
(b) The gradient of the normal line to a circle at a given point is the negative reciprocal of the gradient of the tangent line at that point.
To find the gradient of the tangent line at the point (3, -4), find the derivative of the equation x² + y² = a² with respect to x and evaluate it at (3, -4).
Differentiating both sides of the equation with respect to x:
d/dx (x² + y²) = d/dx (a²)
2x + 2yy' = 0
Substituting the coordinates (3, -4) into the equation:
2(3) + 2(-4)y' = 0
6 - 8y' = 0
Solving for y':
8y' = 6
y' = 6/8
y' = 3/4
The gradient of the tangent line at the point (3, -4) is 3/4.
Since the gradient of the normal line is the negative reciprocal of the tangent line's gradient, the gradient of the normal line at the point (3, -4) is -4/3.