Two circles radii 3cm and rcm touch each other externally at T. if the length of their common tangent PT is 8cm.calculate the value of r
Draw a diagram, and you can see that
(r+3)^2 + (r-3)^2 = 8^2
Well, it seems we have a mathematical circus going on here. Let's bring out the clowns, shall we?
Now, let's get down to business. We have two circles: one with a radius of 3cm, and the other with a radius of rcm. They touch each other externally at point T. We're told that the length of their common tangent PT is 8cm.
Now, if you look closely, you'll notice that we have quite the triangle going on, with sides of length 3cm, rcm, and 8cm. Ah, geometry, you cheeky devil!
We can use the Pythagorean theorem to find the value of r. According to this theorem, the square of the hypotenuse (in this case, 8cm) is equal to the sum of the squares of the other two sides (3cm and rcm). In equation form, it would look a little something like this:
8^2 = 3^2 + r^2
Simplifying this equation, we get:
64 = 9 + r^2
Subtracting 9 from both sides, we have:
55 = r^2
Taking the square root of both sides, we find:
r ≈ √55
So, the value of r is approximately the square root of 55.
And there you have it! The value of r in this delightful little circus act is approximately √55. Enjoy the show!
To find the value of r, we can use the Pythagorean theorem.
Let's denote the center of the larger circle as O and the center of the smaller circle as O'. Also, let's denote the point where the common tangent PT touches the larger circle as A and the point where it touches the smaller circle as B.
Since PT is the common tangent to both circles, it is perpendicular to the line joining the centers OO'. Thus, triangle POO' is a right triangle.
We are given that the length of the common tangent PT is 8 cm. Since PT is perpendicular to OO', it divides OO' into two segments: OP and O'P.
Let's assign new variables:
OP = h
O'P = k
Using Pythagorean theorem for triangle POO', we have:
OO'² = OP² + O'P²
(3+ r)² = h² + k² ...(1)
We are also given that the length of the common tangent PT is 8 cm. Since PT touches the larger circle at A and the smaller circle at B, we can write:
PA + AB + BT = PT
(3 + r) + AB + (3 - r) = 8
6 + AB = 8
AB = 8 - 6
AB = 2
Since PA and BT are radii of the respective circles, PA = 3 and BT = r.
Using the fact that PA and AB are perpendicular, we can apply Pythagorean theorem for triangle PAB:
PA² + AB² = PB²
3² + 2² = (3 + r)²
9 + 4 = 9 + 6r + r²
13 = 9 + 6r + r²
0 = r² + 6r - 4 ...(2)
Now, we have a system of equations (1) and (2). We can solve them to find the value of r.
From equation (1), we have:
(3 + r)² = h² + k²
9 + 6r + r² = h² + k² ...(3)
Substituting (2) into (3), we have:
9 + 6r + r² = h² + k²
9 + 6r + r² = (r² + 6r - 4) + k²
9 + 6r + r² = r² + 6r - 4 + k²
9 = -4 + k²
k² = 13
Since k is a length, we take the positive square root:
k = √13
Therefore, the value of r is √13 cm.
To find the value of r, we can use the properties of tangent circles.
Let's denote the centers of the two circles as O1 and O2, where O1 is the center of the circle with a radius of 3cm, and O2 is the center of the circle with an unknown radius r. Also, let's label the point of tangency as T, and the point where the common tangent PT intersects the line connecting the two centers as M.
We know that the tangent to a circle is perpendicular to the radius at the point of tangency. Therefore, OT1T and OT2T are right triangles.
Since the length of the common tangent PT is 8cm, we can express the lengths in terms of PT.
In triangle OT1T, OT1 is the radius with a length of 3cm, and PT is the hypotenuse with a length of 8cm. So, by using the Pythagorean theorem, we can find the length of OT2.
OT1^2 + PT^2 = OT2^2
3^2 + 8^2 = OT2^2
9 + 64 = OT2^2
73 = OT2^2
Therefore, OT2 = √73 cm.
Now, we have a right-angled triangle OT2T, where OT2 = √73 cm, TT2 = r, and PT = 8 cm.
Applying the Pythagorean theorem to this triangle, we can find the value of r.
OT2^2 = TT2^2 + PT^2
(√73)^2 = r^2 + 8^2
73 = r^2 + 64
9 = r^2
Taking the square root of both sides, we find that r = 3 cm.
So, the value of r is 3 cm.