Question

Find the equation(s) of the tangent(s) to the circle :2.1.with the centre at the origin and passing through the point (5;-3),a point on the circle

Answer 1

since the center is at the origin,

The equation of a circle is given as;

`x^2+y^2=r^2`

Since its passing through the point (5, -3)

`\Rightarrow r=√(5^2+(-3)^2)=√(34)`
`\Rightarrow x^2+y^2=34`

differentiating both sides with respect to x

`\begin{gathered} \Rightarrow2x+2y(dy)/(dx)=0 \\ \\ \Rightarrow(dy)/(dx)=-(x)/(y) \end{gathered}`

At point (5, -3)

The slope m = dy/dx at (5, -3)

`m=-(5)/(-3)=(5)/(3)`

Hence the equation of tangent is given as;

`\begin{gathered} (y-y_1)/(x-x_1)=m \\ \\ \Rightarrow(y+3)/(x-5)=(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=(5)/(3)x-(34)/(3) \end{gathered}`

Hence, the equation of tangent is y = 5/3x -34/3