Answer:
[tex]x-2y+3=0[/tex]
Step-by-step explanation:
Equation of a circle: [tex](x-h)^2+(y-k)^2=r^2[/tex]
(where (h, k) is the centre, and r is the radius)
Given equation: [tex](x-3)^2+(y+2)^2=20[/tex]
Therefore, the centre is (3, -2) and the radius is √20
To find the equation of the tangent
The tangent of a circle is perpendicular to the radius at the point.
Therefore, first find the gradient (slope) of the line that passes through the centre of the circle and the given point.
[tex]\textsf{let}\:(x_1,y_1)=(1,2)[/tex]
[tex]\textsf{let}\:(x_2,y_2)=(3,-2)[/tex]
[tex]\textsf{slope}\:(m)=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{-2-2}{3-1}=-2[/tex]
If two lines are perpendicular to each other, the product of their gradients will be -1.
Therefore, the gradient of the tangent is:
[tex]m_t \cdot-2=-1\implies m_t=\dfrac12[/tex]
Now use the point-slope formula of a linear equation to find the equation of the tangent:
[tex]\implies y-y_1=m_t(x-x_1)[/tex]
[tex]\implies y-2=\dfrac12(x-1)[/tex]
[tex]\implies 2y-4=x-1[/tex]
[tex]\implies x-2y+3=0[/tex]
