Answer to Question #293103 in Analytic Geometry for kyumi

Question

Circles has radius 5 and tangent to the line 3x+4y=24 at the point (2, 412⁄). Illustrate thenfindthe equations of thecircles

Answer 1

Solution:

$r=5$

Circle is tangent to the line 3x+4y=24 at (2, 41/2)

Or tangent is 3x+4y=24.

Slope of tangent $=-\dfrac 34$

Let the equation of the circle be $(x-h)^2+(y-k)^2=r^2$

$\Rightarrow (x-h)^2+(y-k)^2=5^2 \\ \Rightarrow (x-h)^2+(y-k)^2=25\ ...(i)$

On differentiating w.r.t $x$,

$2(x-h)+2(y-k)y'=0 \\\Rightarrow y'=\dfrac{x-h}{y-k}\ \ \ \ \ [Slope\ of\ tangent]$

For (2, 41/2), put $x=2, y=41/2, y'=-3/4$ in above equation.

$-\dfrac 34=\dfrac{2-h}{\frac {41}2-k} \\ \Rightarrow h=-\dfrac{4k-88}{3}$

There is no other info in the problem, as the question is incomplete.

Assume h=0, then k=22

Now put it in (i), we will get the equation of circle.

$x^2+(y-22)^2=25$