Help find eqn of circle given another circle that is tangent

[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:

\(\displaystyle y=\frac{4}{3}x\)

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

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

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

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

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

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

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

\(\displaystyle h=3\pm6\)

This implies:

\(\displaystyle h=-3,\,9\)

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

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

 

[CellCoree]

As a scientist, my first step would be to verify the given information and equations to ensure accuracy. Assuming that the given information is correct, here is my response:

To find the equation of a circle that is tangent to another circle, we can use the fact that the point of tangency lies on the line connecting the centers of the two circles. In this case, the center of the smaller circle is (3,4) and the center of the larger circle is (0,0).

Using the distance formula, we can calculate the distance between these two centers:

d = √[(0-3)^2 + (0-4)^2] = 5

Since the radius of the smaller circle is 10, the distance between the centers must be 10 as well. This means that the point of tangency must be 5 units away from the center of the smaller circle in the direction of the larger circle's center.

To find this point, we can use the equation of a line in point-slope form:

y - y1 = m(x - x1)

Where m is the slope and (x1, y1) is a point on the line. In this case, the slope is given as 4/3 and the point (x1, y1) can be any point on the line connecting the two centers. Let's choose (0,0) as our point.

So, the equation of the line is y - 0 = (4/3)(x - 0) or y = 4/3x.

Now, we can substitute this into the equation of the larger circle, X^2 + y^2 = 25, to find the x-coordinate of the point of tangency:

x^2 + (4/3x)^2 = 25

x^2 + 16x^2/9 = 25

25x^2/9 = 25

x^2 = 9

x = ± 3

Since the point of tangency must be 5 units away from the center of the smaller circle in the direction of the larger circle's center, we can conclude that the x-coordinate of the point of tangency is 3.

Now, we can substitute this into the equation of the line to find the y-coordinate:

y = 4/3(3) = 4

Therefore, the point of tangency is (3,4).