Answer:
Option C) is correct
Step-by-step explanation:
Given: Endpoints of the diameter of the circle are A(-1, -2) and B(3,10)
To find: slope of the tangent drawn to the circle at point B
Solution:
Let [tex](x_1,y_1)=(-1,-2)\,,\,(x_2,y_2)=(3,10)[/tex]
Centre of the circle = [tex](\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})=(\frac{-1+3}{2},\frac{-2+10}{2})=(1,4)[/tex]
Let [tex](h,k)=(1,4)[/tex]
Distance formula states that distance between points (a,b) and (c,d) is given by [tex]\sqrt{(c-a)^2+(d-b)^2}[/tex]
Radius of the circle = Distance between points [tex](-1,-2)[/tex] and [tex](1,4)[/tex] = [tex]\sqrt{(1+1)^2+(4+2)^2}=\sqrt{4+36}=\sqrt{40}[/tex] units
Let r = [tex]\sqrt{40}[/tex] units
Equation of a circle is given by [tex](x-h)^2+(y-k)^2=r^2[/tex]
[tex](x-1)^2+(y-4)^2=\left ( \sqrt{40} \right )^2\\(x-1)^2+(y-4)^2=40[/tex]
Differentiate with respect to x
[tex]2(x-1)+2(y-4)\frac{\mathrm{d} y}{\mathrm{d} x}=0\\\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1-x}{y-4}[/tex]
Put [tex](x,y)=(3,10)[/tex]
[tex]\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1-3}{10-4}=\frac{-2}{6}=\frac{-1}{3}[/tex]
So,
slope of the tangent drawn to this circle at point B = [tex]\frac{-1}{3}[/tex]
