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Relation between Diameter and Perimeter

A and B are...

Question

$A$ and $B$ are centres of circles of radii $9 c m$ and $2 c m$ such that $A B = 17 c m$ and $C$ is the centre of the circle of radius $r c m$ which touches the above circles externally. If $∠ A C B = 90^{o}$ , The equation for $r$ and the value of r are respectively

r^{2} + 11 r - 102 = 0

;

r = 17 c m

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r^{2} + 11 r - 204 = 0

;

r = 6 c m

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r^{2} + 11 r - 102 = 0

;

r = 6 c m

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r^{2} + 22 r - 102 = 0

;

r = 17 c m

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Solution

The correct option is C $r^{2} + 11 r - 102 = 0$ ; $r = 6 c m$
$A C = 9 + r c m$
$A B = 17 c m$
$B C = r + 2 c m$
$A B^{2} = A C^{2} + B C^{2}$
$\Rightarrow 17^{2} = (r + 9)^{2} + (r + 2)^{2}$
$\Rightarrow r^{2} + 81 + 18 r + r^{2} + 4 + 4 r = 289$
$\Rightarrow 2 r^{2} + 22 r - 204 = 0$
$\Rightarrow r^{2} + 11 r - 102 = 0$
$\Rightarrow r^{2} + 17 r - 6 r - 102 = 0$
$\Rightarrow r (r + 17) - 6 (r + 17) = 0$
$\Rightarrow (r + 17) (r - 6) = 0$
$\Rightarrow r = 6 c m$
$∴ C)$ Answer.

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Similar questions

A

B

are centres of circles of radii

9

cm and

2

cm respectively.

A B = 17 c m

R

is the center of the circle of radius

x c m

which touches the above circle externally. Given that angle

A R B

90^{o}

. Write an equation in

x

and solve it.

A

: lf the circles

x^{2} + y^{2} = a^{2},

x^{2} + y^{2} - 6 x - 8 y + 9 = 0

touch externally then

a = 1

R

: Two circles with centres

C_{1}, C_{2}

and radii

r_{1}, r_{2}

respectively touch externally iff

C_{1} C_{2} = r_{1} + r_{2}

Q. Let

A B C

be a triangle with in centre

I

and in radius

r

. Let

D, E, F

be the feet of the perpendiculars from

I

to the sides

B C, C A

and

A B

respectively. If

r_{1}, r_{2} r_{3}

are the radii of circles inscribed in then prove

\frac{r_{1}}{r - r_{1}}_\frac{r_{2}}{r - r_{2}} + \frac{r_{3}}{r - r_{3}} = \frac{r_{1} r_{2} r_{3}}{(r - r_{1}) (r - r_{2}) (r - r_{3})}

Q. Two circles with centre

A a n d B

have radii

8 c m a n d 1 c m

respectively. The distance

A B

13 c m

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C

and radius

r c m

touches both circles externally. If angle

A B C

90^{\circ}

, then value of

r

Q. If the sum of the areas of two circles with radii R₁ and R₂ is equal to the area of a circle of radius R, then
(a)

R_{1} + R_{2} = R

(b)

R_{1} + R_{2} < R

(c)

R_{1}^{2} + R_{2}^{2} < R^{2}

(d)

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A and B are centres of circles of radii 9cm and 2cm such that AB=17cm and C is the centre of the circle of radius rcm which touches the above circles externally. If ∠ACB=90o, The equation for r and the value of r are respectively

The correct option is C r2+11r−102=0; r=6cmAC=9+rcmAB=17cmBC=r+2cmAB2=AC2+BC2⇒172=(r+9)2+(r+2)2⇒r2+81+18r+r2+4+4r=289⇒2r2+22r−204=0⇒r2+11r−102=0⇒r2+17r−6r−102=0⇒r(r+17)−6(r+17)=0⇒(r+17)(r−6)=0⇒r=6cm∴C) Answer.

$A$ and $B$ are centres of circles of radii $9 c m$ and $2 c m$ such that $A B = 17 c m$ and $C$ is the centre of the circle of radius $r c m$ which touches the above circles externally. If $∠ A C B = 90^{o}$ , The equation for $r$ and the value of r are respectively