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Circles with radius of 6 units and 4 units are drawn in such a way that exactly 3 common tangents are possible. If a direct common tangent from point A touches the circles at points B and C, what is the length of BC?
- a)8√6
- b)7√6
- c)4√6
- d)12
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
Circles with radius of 6 units and 4 units are drawn in such a way tha...
Let O1 and O2 be the centers of the circles with radii 6 and 4 units, respectively. Let D be the point where the direct common tangent from A touches the larger circle, and let E be the point where the direct common tangent from A touches the smaller circle.
Since the radius of the larger circle is 6 units, OD is perpendicular to the direct common tangent from A. Similarly, OE is perpendicular to the direct common tangent from A.
Since the direct common tangent from A is a tangent to both circles, BD and CE are also tangents to the circles. Therefore, BD is perpendicular to OD, and CE is perpendicular to OE.
Let F be the intersection of BD and CE. Since BD is perpendicular to OD and CE is perpendicular to OE, BD and CE are parallel. Therefore, triangle BCF is a right triangle.
Since the radius of the larger circle is 6 units and OD is perpendicular to the direct common tangent from A, OD has length 6 units. Similarly, OE has length 4 units.
Since triangle BCF is a right triangle, BF is the hypotenuse. By the Pythagorean theorem, we have:
BF^2 = BC^2 + CF^2
Since triangle BCF is a right triangle, CF has length 6 - 4 = 2 units.
Substituting the values we know into the equation above, we have:
BF^2 = BC^2 + 2^2
BF^2 = BC^2 + 4
Since BF is the hypotenuse of a right triangle, its length is always greater than the lengths of the other two sides. Therefore, BF > BC.
Since BC and BF are both positive lengths, we can take the square root of both sides of the equation above:
sqrt(BF^2) = sqrt(BC^2 + 4)
BF = sqrt(BC^2 + 4)
Since BF > BC, we have:
sqrt(BC^2 + 4) > BC
Squaring both sides of this inequality, we have:
BC^2 + 4 > BC^2
4 > 0
Therefore, the inequality is always true. This means that any positive value of BC satisfies the inequality.
Therefore, the length of BC can be any positive value. It is not necessarily 8 units.
Since the radius of the larger circle is 6 units, OD is perpendicular to the direct common tangent from A. Similarly, OE is perpendicular to the direct common tangent from A.
Since the direct common tangent from A is a tangent to both circles, BD and CE are also tangents to the circles. Therefore, BD is perpendicular to OD, and CE is perpendicular to OE.
Let F be the intersection of BD and CE. Since BD is perpendicular to OD and CE is perpendicular to OE, BD and CE are parallel. Therefore, triangle BCF is a right triangle.
Since the radius of the larger circle is 6 units and OD is perpendicular to the direct common tangent from A, OD has length 6 units. Similarly, OE has length 4 units.
Since triangle BCF is a right triangle, BF is the hypotenuse. By the Pythagorean theorem, we have:
BF^2 = BC^2 + CF^2
Since triangle BCF is a right triangle, CF has length 6 - 4 = 2 units.
Substituting the values we know into the equation above, we have:
BF^2 = BC^2 + 2^2
BF^2 = BC^2 + 4
Since BF is the hypotenuse of a right triangle, its length is always greater than the lengths of the other two sides. Therefore, BF > BC.
Since BC and BF are both positive lengths, we can take the square root of both sides of the equation above:
sqrt(BF^2) = sqrt(BC^2 + 4)
BF = sqrt(BC^2 + 4)
Since BF > BC, we have:
sqrt(BC^2 + 4) > BC
Squaring both sides of this inequality, we have:
BC^2 + 4 > BC^2
4 > 0
Therefore, the inequality is always true. This means that any positive value of BC satisfies the inequality.
Therefore, the length of BC can be any positive value. It is not necessarily 8 units.
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Community Answer
Circles with radius of 6 units and 4 units are drawn in such a way tha...
Since exactly 3 tangents are possible, the circles should touch externally. Let us try to draw the circles and a direct common tangent to understand the question better.
Here O'C = 6 and OB = 4. BC is the direct common tangent and D is the transverse tangent. We have to find length BC.
Triangle BOA and CO'A are similar triangles.
Let OA = x then
On cross-multiplying we get 6x = 40 + 4x thus x = 20
Thus O'A = 6+4+20=30
In right triangle O'AC
(AC)2=(O′A)2−(O′C)2
(AC)2=(30)2−(6)2=864
(AC)=12√6....(I)
In right triangle OAB
(AB)2=(OA)2−(OB)2
(AB)2=(20)2−(4)2=384
(AB)=8√6....(II)
BC = AC - AB From(I) and (II)
BC = 4√6
Triangle BOA and CO'A are similar triangles.
Let OA = x then
On cross-multiplying we get 6x = 40 + 4x thus x = 20
Thus O'A = 6+4+20=30
In right triangle O'AC
(AC)2=(O′A)2−(O′C)2
(AC)2=(30)2−(6)2=864
(AC)=12√6....(I)
In right triangle OAB
(AB)2=(OA)2−(OB)2
(AB)2=(20)2−(4)2=384
(AB)=8√6....(II)
BC = AC - AB From(I) and (II)
BC = 4√6
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Circles with radius of 6 units and 4 units are drawn in such a way that exactly 3 common tangents are possible. If a direct common tangent from point A touches the circles at points B and C, what is the length of BC?a)8√6b)7√6c)4√6d)12Correct answer is option 'C'. Can you explain this answer?
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Circles with radius of 6 units and 4 units are drawn in such a way that exactly 3 common tangents are possible. If a direct common tangent from point A touches the circles at points B and C, what is the length of BC?a)8√6b)7√6c)4√6d)12Correct answer is option 'C'. Can you explain this answer? for CAT 2024 is part of CAT preparation. The Question and answers have been prepared according to the CAT exam syllabus. Information about Circles with radius of 6 units and 4 units are drawn in such a way that exactly 3 common tangents are possible. If a direct common tangent from point A touches the circles at points B and C, what is the length of BC?a)8√6b)7√6c)4√6d)12Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for CAT 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Circles with radius of 6 units and 4 units are drawn in such a way that exactly 3 common tangents are possible. If a direct common tangent from point A touches the circles at points B and C, what is the length of BC?a)8√6b)7√6c)4√6d)12Correct answer is option 'C'. Can you explain this answer?.
Circles with radius of 6 units and 4 units are drawn in such a way that exactly 3 common tangents are possible. If a direct common tangent from point A touches the circles at points B and C, what is the length of BC?a)8√6b)7√6c)4√6d)12Correct answer is option 'C'. Can you explain this answer? for CAT 2024 is part of CAT preparation. The Question and answers have been prepared according to the CAT exam syllabus. Information about Circles with radius of 6 units and 4 units are drawn in such a way that exactly 3 common tangents are possible. If a direct common tangent from point A touches the circles at points B and C, what is the length of BC?a)8√6b)7√6c)4√6d)12Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for CAT 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Circles with radius of 6 units and 4 units are drawn in such a way that exactly 3 common tangents are possible. If a direct common tangent from point A touches the circles at points B and C, what is the length of BC?a)8√6b)7√6c)4√6d)12Correct answer is option 'C'. Can you explain this answer?.
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