Finding circles with radius 10 tangent to x² + y² = 25 at (3,4)

[sktrinh]

please help me find the standard equation of the circles that have radius 10 and are tangent to the circle X^2 + y^2 = 25 at the point (3,4).

the soln: (x-9)^2 + (y-12)^2 = 100, (x+3)^2 + (y+4)^2 = 100,

i found the eqn that intersects the centre of the small circle and the larger one to be: y=4/3x, substituted as k=4/3k for C(h,k) into the eqn of the circle, however require some help solving it. thanks!

 

[MarkFL]

I would simply observe that the center of one circle must be at (-3,-4) and the other at (9,12). These points are both 10 units from the tangent point along the line you correctly found.

 

[sktrinh]

I would simply observe that the center of one circle must be at (-3,-4) and the other at (9,12). These points are both 10 units from the tangent point along the line you correctly found.

is there a algebraic way of solving it using distance formulas like d = |ax+by+c|/sqrt(a^2+b^2°+) or something else?

I solved for eqn of line of the that passes thru both the small and large circle being y=4/3x, and set k=4/3h since it passes thru the large circle as well (i think), with this expression i plugged into (3-h)^2 + (4-k)^2 = 100

(3-h)^2 + (4-4/3h)^2 = 100

I couldn't solve this thru, i keep getting h^2 -6h -9, which i think shud be h^2 -6h + 9, so that h = -3 and which would follow k = -4. How would I solve in a similar fashion for the second large circle as impled by the solution of C(9,12)??

 

[MarkFL]

Well, we know the centers of all circles will lie on the line:

$$y=\frac{4}{3}x$$

And so the center of the two tangent circles will be at:

$$(h,k)=\left(h,\frac{4}{3}h\right)$$

And since the must both pass through the point $(3,4)$, we may state:

$$(3-h)^2+\left(4-\frac{4}{3}h\right)^2=100$$

$$(3-h)^2+\frac{16}{9}\left(3-h\right)^2=100$$

$$\frac{25}{9}\left(3-h\right)^2=100$$

$$\left(h-3\right)^2=6^2$$

$$h=3\pm6$$

This implies:

$$h=-3,\,9$$

And using the relation between $h$ and $k$, which is $$k=\frac{4}{3}h$$ , we then conclude the centers are at:

$$(-3,-4),\,(9,12)$$