1) Find the equation of a line(circle?) through the origin with centre at the point(4,-7).
distance (0,0) to (4,-7) This will be the radius.
\(d=\sqrt{16+49}=\sqrt{65}\)
\((x-4)^2+(y+7)^2=65\)
Find the coordinates of the point which this circles meets the line y=1.
\((x-4)^2+(1+7)^2=65\\ (x-4)^2+64=65\\ (x-4)^2=1\\ x-4=\pm 1\\ x=5,\;\;\;or\;\;\;x=3\\ (5,1) \quad and \quad (3,1)\)
Find also the equation of the tangent at this point,
What point??
other than the origin, where the circle meet the axis.
\((x-4)^2+(y+7)^2=65\)
If x=0
\((0-4)^2+(y+7)^2=65\\ 16+(y+7)^2=65\\ (y+7)^2=49\\ y+7=\pm7\\ y=0 \quad or \quad -14\)
If y=0
\((x-4)^2+(0+7)^2=65\\ (x-4)^2=16\\ x-4=\pm4\\ x=0 \quad or \quad 8\)
So the y intercepts are 0 and -14
and the x intercepts are 0 and 8
