Solution
since the center is at the origin,
The equation of a circle is given as;
[tex]x^2+y^2=r^2[/tex]Since its passing through the point (5, -3)
[tex]\Rightarrow r=\sqrt{5^2+(-3)^2}=\sqrt{34}[/tex][tex]\Rightarrow x^2+y^2=34[/tex]differentiating both sides with respect to x
[tex]\begin{gathered} \Rightarrow2x+2y\frac{dy}{dx}=0 \\ \\ \Rightarrow\frac{dy}{dx}=-\frac{x}{y} \end{gathered}[/tex]At point (5, -3)
The slope m = dy/dx at (5, -3)
[tex]m=-\frac{5}{-3}=\frac{5}{3}[/tex]Hence the equation of tangent is given as;
[tex]\begin{gathered} \frac{y-y_1}{x-x_1}=m \\ \\ \Rightarrow\frac{y+3}{x-5}=\frac{5}{3} \\ \\ \text{ If we cross multply} \\ \\ \Rightarrow3(y+3)=5(x-5) \\ \\ \\ \Rightarrow3y+9=5x-25 \\ \\ \Rightarrow3y=5x-25-9 \\ \\ \Rightarrow3y=5x-34 \\ \\ \Rightarrow y=\frac{5}{3}x-\frac{34}{3} \end{gathered}[/tex]Hence, the equation of tangent is y = 5/3x -34/3
