Length of internal (transverse) common tangent to two circles is Lint=√d2−(r1−r2)2
Where d = distance between the centers of the two circles r1& r2 are radii of two circles
False
Let center of two circles being c1& c2 and their radii r1& r2
Then,
Given,
c1 c2 = d
Let transverse tangent touches the circle at A and B. It intersects the line segment c1 c2 at P.
Let length of transverse tangent AB = Ltrans
Right angled triangle c1 AP
C1P2 = r22+ AP2- - - - - - (1)
Right angles triangle C2BP
C2P2 = r22+ PB2- - - - - - (2)
In △ AC1P & △ BC2P
∠ C1AP = ∠ C2BP = 90∘
∠ APc1 =∠ BPc2 = oppositeangle
Since, two angles of a triangle are equal, third angle should also be equal. {sum of the angle of triangle = 180∘
△ AC1P ≅ △ BC2P {AAA similarity}
Ac1Bc2=ApBp=c1pc2p
r2r1=Ltrans−BPBP=d−C2PC2P→______←r2r1=Ltransr1r1+r2r2r1=d−C2PC2P C2P=dr1r1+r2
Substituting the values of BP & c2P in equation (2)
d2r12(r1+r2)2=r12+L2transr12(r1+r2)2
d2=(r1+r2)2+L2trans
L2trans=d2−(r1+r2)2
Ltrans=√d2−(r1+r2)2
So,given statment
Length of internal (transverse) common tangent to circles is
Lint=√d2−(r1−r2)2 is NOT correct.