Question

Two circles with radii a and b respectively touch each other externally. Let c be the radius of a circle that touches these two circles as well as a common tangent to these two circles. Then_____

Need solution for brainlist​

Answer 1

Answer:

Answer. Given X and Y are two circles touch each other externally at P. AB is the common tangent to the circles X and Y at point A and B respectively.

To find : ∠APB

Proof: let ∠CAP = α and ∠CBP = β.

CA = CP [lengths of the tangents from an external point C]

In a triangle PAC, ∠CAP = ∠APC = α

similarly CB = CP and ∠CPB = ∠PBC = β

now in the triangle APB,

∠PAB + ∠PBA + ∠APB = 180° [sum of the interior angles in a triangle]

α + β + (α + β) = 180°

2α + 2β = 180°

α + β = 90°

∴ ∠APB = α + β = 90°

Answer 2

Answer:

Then : (1/√c) = (1/√a) + (1/√b)

Step-by-step explanation:

From Figure,

PR = MC = √AC² - AM²

⇒ √(a + c)² - (a - c)² = 2√ac

⇒ QR = 2√bc

Now,

PQ = PR = RQ = 2√ac + 2√bc   ---- (1)

Draw PN ║ AB.

∴ PN = AB = a + b

⇒ QN = BQ - BN

          = b - a

∴ PQ² = PN² - QN²

⇒ PQ² = (a + b)² - (a - b)²

⇒ PQ² = 4ab

⇒ PQ = 2√ab    ----- (2)

From (1) and (2), we get

(1/√c) = (1/√a) + (1/√b)

Hope it helps!