Question

X and Y are centers of circles of radius 9cm and 2cm and XY = 17cm. Z is the center of a circle of radius 4cm, which touches the above circles externally. Given that $\angle$ = 90°, Write an equation in r and solve it for r.​

Answer 1

$\huge\underline\mathbb{SOLUTION:-}$

$\mathsf {Let\:r\:be\:the\:radius\:of\:the\:third\:circle}$

$\mathsf {XY = 17\:Cm \implies XZ = 9 + r\:YZ = 2}$

$\underline\mathsf {ATQ:-}$

$\mathsf {(r + 9)^2 + (r + 2)^2 = (1 + r)^2}$

$\implies \mathsf {r^2 + 18r + 81 + r^2 + 4r + 4 = 289S}$

$\implies \mathsf {r^2 + 11r - 10r = 0}$

$\mathsf {(r + 17) (r - 6) = 0}$

$\implies \mathsf {r = -17\:(N.P)}$

$\implies \mathsf {r = 6\:cm}$

$\therefore \mathsf \blue {Radius = 6\:Cm}$

Answer 2

Answer:

SOLUTION:−

\mathsf {Let\:r\:be\:the\:radius\:of\:the\:third\:circle}Letrbetheradiusofthethirdcircle

\mathsf {XY = 17\:Cm \implies XZ = 9 + r\:YZ = 2}XY=17Cm⟹XZ=9+rYZ=2

\underline\mathsf {ATQ:-}

ATQ:−

\mathsf {(r + 9)^2 + (r + 2)^2 = (1 + r)^2}(r+9)

2

+(r+2)

2

=(1+r)

2

\implies \mathsf {r^2 + 18r + 81 + r^2 + 4r + 4 = 289S}⟹r

2

+18r+81+r

2

+4r+4=289S

\implies \mathsf {r^2 + 11r - 10r = 0}⟹r

2

+11r−10r=0

\mathsf {(r + 17) (r - 6) = 0}(r+17)(r−6)=0

\implies \mathsf {r = -17\:(N.P)}⟹r=−17(N.P)

\implies \mathsf {r = 6\:cm}⟹r=6cm

\therefore \mathsf \blue {Radius = 6\:Cm}∴Radius=6Cm

Step-by-step explanation:

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