Question

A circle with area (25/9)*pi touches the x-axis at the point (4,0).The point T is the furthest point on the circle from the origin O.Find the length of OT giving your answer as a simplified fraction.

Answer 1

Answer on Question #53273 – Math – Analytic Geometry

Question:

A circle with area $(25/9)*pi$ touches the $x$-axis at the point $(4,0)$.

The point $T$ is the furthest point on the circle from the origin $O$. Find the length of $OT$ giving your answer as a simplified fraction.

Solution

If circle's area is $(25/9)*pi$, the formula for area of circle is $S = pi * r^2$, where $r$ is the length of circle's radius, then

$$(25/9)*pi = pi * r^2, \text{ so } r = 5/3.$$

If circle's radius is $5/3$ and circle touches the $x$-axis at point $(4,0)$, then the equation of the circle is $(x-4)^2 + (y-5/3)^2 = 25/9$. The center of the circle is $(4, 5/3)$.

furtherst lie on a diameter

$$y = mx + c$$

gradient $m = y/x = (5/3)/4 = 5/12$

$$y = mx + c$$

$5/3 = (5/12)*4 + c$. so $c = 0$.

The line from the center of the circle to the origin is $y = 5x/12$. Find the intersections of line and circle.

$$(x-4)^2 + ((5x/12) - 5/3)^2 = 25/9$$ $$x^2 - 8x + 16 + 25x^2 / 144 + 25/9 - 25/18 = 25/9$$ $$x = 32/13; 72/13.$$

farthest from origin: $(72/13, 30/13)$.

So the coordinates of $T$ is $(72/13, 30/13)$.

the length of OT is: $\mathrm{OT} = ((72/13)^2 + (30/13)^2)^{1/2} = 6$.

Answer: $\mathrm{OT} = 6$

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