Currently I am looking at a graph of a circle. The diameter is y=2x+3 Tangent at point E cuts the x-axis at F (12;0) 1. find the coordinates of E 2. find the coordinates of G and H (H being the centre)
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Hints: The diameter lies on the line $y=2x+3$, which has a slope of $2$, so the tangent line has a slope of $-\frac12$. The tangent line contains the point $(12,0)$, so using the point-slope form we get $$ y-0=-\frac12(x-12) $$ as the equation of the tangent line.
(1) The point $E$ is the intersection of the tangent line with the diameter line.
(2) $G$ is the $y$-intercept of the tangent line.
(3) Suppose the centre $H$ of the circle has coordinates $(a,b)$. The centre $H$ lies on the diameter line, so $a$ and $b$ satisfy $$ b=2a+3.$$ That's one equation for $a$ and $b$. But additional information is needed to completely determine the values of $a$ and $b$. For example, it looks like $H$ lies on the $y$-axis; if so, then this implies that $a=0$, and you can solve for $b$. Without additional information you can imagine infinitely many possibilities for the location of $H$.
(4) The shape $OFEH$ can be viewed as the difference of two triangles, $OFG$ and $HEG$.