1) Find the equation of a line through the origin with centre at the point(4,-7). Find the coordinates of the point which this circles meets the line y=1. Find also the equation of the tangent at this point, other than the origin, where the circle meet the axis.
2) Find the coordinates from the foot of the perpendicular from the point(1,2) to the line 4x-3y+7=0 and prove that this touches the circle x2 +y2 -2x-4y+4=0.
+118654 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
+118654 2) Find the coordinates from the foot of the perpendicular from the point(1,2) to the line 4x-3y+7=0
I assume this means "Where does the perpendicular from (1,2) to 4x-3y+7=0 intersect 4x-3y+7=0"
Gradient of 4x-3x+7=0 is 4/3
Gradient of the perpendicular is -3/4
\(y=\frac{-3}{4}x+k\qquad sub\;\;in\;\; (1,2)\\ 2=\frac{-3}{4}*1+k\\ 2\frac{3}{4}=k\\ k=\frac{11}{4}\\ y=\frac{-3x+11}{4}\\ 4y+3x-11=0 \)
and prove that this touches the circle x2 +y2 -2x-4y+4=0.
\(x^2 +y^2 -2x-4y+4=0\\ (x^2-2x) +(y^2 -4y)=-4\\ (x^2-2x+1) +(y^2 -4y+4)=-4+1+4\\ (x-1)^2 +(y-2)^2=1\\ \text{This is a circle with centre (1,2) and radius 1}\)
Since (1,2) is the centre of the circle, the line passes right through the middle of the the circle so it must intersect with it twice.
