two circles with centres A and B of radius 3 cm and 4 cm respectively intersect at two points C and D such that AC and BC are tangents to the two circles. find the length of the common chord CD
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HELLO DEAR,GIVEN THAT:-
AC=3cm
,BC=4cm
IN∆ACB,<ACB=90°
(AB)²=(AC)²+(BC)²
(AB)²=9+16
(AB)=5cm
IN ∆ACP, <APC=90°
LET (CP)=Xcm
AND,
AP=Ycm
(CP)² +(AP)²= (AC)²
x²+y²=3²
x² = (9-y²)....-----------------(1)
IN∆BPC, <BPC=90°
BP=(5-y)
X²+(5-y)²=4²
x²=16-(5-y)²
x³=16-25-y²+10y
9-y²=-9-y²+10y
-y²+y²+9+9=10y
10y=18
y=1.8--------------(2)
x²={9-(1.8)²}
x²=9-3.24
x²=5.76
x=2.4cm
LENTH OF THE CHORDS CD=2×CP
=> 2×X.
CD=2×2.4 =4.8CM
I HOPE ITS HELP YOU DEAR,
THANKS