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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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