Question

Construct a tangent to a circle of radius 4cm from a point on the concentric circle of radius 6cm and measure it's length. Also verify the measurement by actual calculation.​

Answer 1

Answer,

$\sf{ (1)Draw \: two \: concentric \: circle \: C(1) \: and \: C(2) \: with \: common \: center}$

$\sf{0 \: and \: radius \: 4cm \: and \: 6cm}$

$\sf{(2) \: Take \: a \: point \: P \: on \: the \: outer \: circle \: C(2) \: and \: join \: OP.}$

$\sf{(3) \: Draw \: the \: bisector \: of \: OP \: which \: bisect \: OP \: at \: M'.}$

$\sf{ (4) \: Taking \: M' \: as \: center \: and \: OM' \: as \: radius \: draw \: a \: dotted \: circle \: which}$

$\sf{cut \: the \: inner \: circle \: C(1) \: at \: two \: point \: M \: and \: P.}$

$\sf{(5) \: Join \: PM \: and \: PP'.  Thus, \: PM \: and \: PP' \: are \: required \: tangent.}$

$\sf{On \: measuring \: PM \: and \: PP'.}$

$\sf{PM=PP′=4.4cm}$

$\sf{By \: calculation:}$

$\sf{In \: ΔOMP,∠PMO=900}$

$\sf{PM {}^{2} =OP {}^{2} −OM {}^{2} (by \: pythagoras \: theorem)}$

$\sf{PM {}^{2} =(6) {}^{2} −(4) {}^{2} }$

$\sf{=36−16=20}$

$\sf{PM {}^{2} =20cm}$

$\sf{PM=20=4.4 \: cm}$

$\sf{ \purple{Hence,the \: length \: of \: the \: tangentis \: 4.4cm}}$

❣Hope this helps you mate ✌