Two Circles and A Triangle

[MathBuster Forum]

Two circles with centers O and P, each with a radius of 2, are tangent to each other. A straight line is drawn through O and P meeting the circles at A and B. Two other sides of triangle ABC are drawn such that side AC is tangent to the circle with center P at D and side CB is tangent to the circle with center P at B.

Determine the length of BC.

Fact: A line drawn from the centre of a circle to the point where the circle meets the tangent is perpendicular to the tangent. 

Fact: If two circles are tangent to each other, a line segment joining the two centres passes through the point of tangency.

[lalitha srinivasan]

BC= twice sq.rt.of 2

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[Dr. Atul Nischal]

Hi Lalitha

Can you share how you for this answer?

A solution is always more interesting than the answer. If you having difficulty typing, you can also click the image of the solution on paper and send it as an image :-)

Cheers

Atul

On Saturday, April 18, 2015 at 7:09:19 AM UTC+5:30, lalitha.srinivasan69 wrote:

BC= twice sq.rt.of 2

On Mon, Apr 13, 2015 at 2:10 AM, MathBuster Forum <MathBusterForum@googlegroups.com> wrote:

Two circles with centers O and P, each with a radius of 2, are tangent to each other. A straight line is drawn through O and P meeting the circles at A and B. Two other sides of triangle ABC are drawn such that side AC is tangent to the circle with center P at D and side CB is tangent to the circle with center P at B.

Determine the length of BC.

Fact: A line drawn from the centre of a circle to the point where the circle meets the tangent is perpendicular to the tangent. 

Fact: If two circles are tangent to each other, a line segment joining the two centres passes through the point of tangency.

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[Writabrat Sharma]

PD is a radius which touches the tangent AC at the point of contact D

Therefore, PD is perpendicular to AC

So, AD = √(AP² – PD²)

ð  AD = √( 6² – 2² )

ð  AD = √(36 – 4)

ð  AD= √(32)

 

ABC is right triangle

So, AC²= AB² + BC²

ð  (AD+DC)² = 8² + BC²

ð  (√32 + DC)² = 64 + DC²   [ bcoz, BC = DC, as they are tangents to the circle from common point]

ð  32 + DC² + 2√32*DC = 64 + DC²

ð  2√32*DC= 64 – 32

ð  2√32*DC = 32

ð  DC = √32/2

ð  DC = 4√2/2 = 2√2.

Finally, its done…..as DC = 2√2 so, BC = DC = 2√2

On Sat, Apr 18, 2015 at 7:37 AM, Dr. Atul Nischal <atu...@elipsis.in> wrote:

Hi Lalitha

Can you share how you for this answer?

A solution is always more interesting than the answer. If you having difficulty typing, you can also click the image of the solution on paper and send it as an image :-)

Cheers

Atul

On Saturday, April 18, 2015 at 7:09:19 AM UTC+5:30, lalitha.srinivasan69 wrote:

BC= twice sq.rt.of 2

On Mon, Apr 13, 2015 at 2:10 AM, MathBuster Forum <MathBus...@googlegroups.com> wrote:

Two circles with centers O and P, each with a radius of 2, are tangent to each other. A straight line is drawn through O and P meeting the circles at A and B. Two other sides of triangle ABC are drawn such that side AC is tangent to the circle with center P at D and side CB is tangent to the circle with center P at B.

Determine the length of BC.

Fact: A line drawn from the centre of a circle to the point where the circle meets the tangent is perpendicular to the tangent. 

Fact: If two circles are tangent to each other, a line segment joining the two centres passes through the point of tangency.

--
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