Solution by Analytic Geometry
$x^2 + y^2 = 25$
$xx_1 + yy_1 = 25$ ← equation of tangent at any point on the curve
At (4, 3), the equation of tangent is
$4x + 3y = 25$
Note:
$Ax + By + C_1 = 0$ and $Bx - Ay + C_2 = 0$ are perpendicular lines.
Hence, the equation of normal through (4, 3) is
$3x - 4y = 3(4) - 4(3)$
$3x - 4y = 0$ ← answer
Solution by Calculus with Analytic Geometry
$x^2 + y^2 = 25$
$2x + 2y \, y' = 0$
$y' = -\dfrac{x}{y}$ ← slope of the tangent at any point
The slope of normal is therefore
$m = \dfrac{y}{x}$
At point (4, 3)
$m = \dfrac{3}{4}$
Equation of normal
$y - y_1 = m(x - x_1)$
$y - 3 = \frac{3}{4}(x - 4)$
$4y - 12 = 3x - 12$
$3x - 4y = 0$ ← answer
Solution by Drawing
$x^2 + y^2 = 5^2$ is a circle with radius 5 units and center at the origin. Note that all lines normal to circle will pass through the center of the circle.
From the figure, the equation of normal through (4, 3) is...
$y = mx + b$
$y = \dfrac{3}{4}x + 0$
$4y = 3x$
$3x - 4y = 0$ ← answer
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