In a circle, two parallel chords on the same side of a diameter have lengths 4 cm and 6 cm. If the distance between these chords is 1 cm, then the radius of the circle, in cm, is
- √12
- √14
- √13
- √11
The Correct Option is C
Solution and Explanation
To find the radius of a circle with two parallel chords of lengths 4 cm and 6 cm, separated by a distance of 1 cm, consider the following approach:
Let the circle's center be O. Assume the first chord of length 6 cm is AB and the second chord of length 4 cm is CD. Both chords are parallel and lie on the same side of the circle's diameter.
Let the perpendicular distance from the center O to the chord AB be d1, and to chord CD be d2. Then:
d2 = d1 + 1 cm.
The distance from the center to a chord is related to the radius and half of the chord length through the Pythagorean theorem:
For chord AB:
\( d1^2 + 3^2 = r^2 \)
(since half of 6 cm is 3 cm).
Thus,
\( d1^2 = r^2 - 9 \) (Equation 1)
For chord CD:
\( d2^2 + 2^2 = r^2 \)
(since half of 4 cm is 2 cm).
Thus,
\( (d1 + 1)^2 + 4 = r^2 \)
Expanding this equation:
\( d1^2 + 2d1 + 1 + 4 = r^2 \)
\( d1^2 + 2d1 + 5 = r^2 \) (Equation 2)
Substitute Equation 1 into Equation 2:
\( (r^2 - 9) + 2d1 + 5 = r^2 \)
\( r^2 - 9 + 2d1 + 5 = r^2 \)
\( 2d1 - 4 = 0 \)
\( 2d1 = 4 \)
\( d1 = 2 \)
Substitute \( d1 = 2 \) back into Equation 1:
\( 2^2 = r^2 - 9 \)
\( 4 = r^2 - 9 \)
\( r^2 = 13 \)
Thus, the radius of the circle is \( \sqrt{13} \) cm.
The correct answer is \( \sqrt{13} \).
Top Questions on Circles
- The absolute difference between the squares of the radii of the two circles passing through the point \( (-9, 4) \) and touching the lines \( x + y = 3 \) and \( x - y = 3 \), is equal to:
- Let $ ABC $ be the triangle such that the equations of lines $ AB $ and $ AC $ are: $ 3y - x = 2 \quad \text{and} \quad x + y = 2, $ respectively, and the points $ B $ and $ C $ lie on the x-axis. If $ P $ is the orthocentre of the triangle $ ABC $, then the area of the triangle $ PBC $ is equal to:
- The given figure shows a circle with centre O and radius 4 cm circumscribed by \(\triangle ABC\). BC touches the circle at D such that BD = 6 cm, DC = 10 cm. Find the length of AE.
- Let $ C_1 $ be the circle in the third quadrant of radius 3, that touches both coordinate axes. Let $ C_2 $ be the circle with center $ (1, 3) $ that touches $ C_1 $ externally at the point $ (\alpha, \beta) $. If $ (\beta - \alpha)^2 = \frac{m}{n} $, and $ \gcd(m, n) = 1 $, then $ m + n $ is equal to:
In the following figure chord MN and chord RS intersect at point D. If RD = 15, DS = 4, MD = 8, find DN by completing the following activity:
Activity :
\(\therefore\) MD \(\times\) DN = \(\boxed{\phantom{SD}}\) \(\times\) DS \(\dots\) (Theorem of internal division of chords)
\(\therefore\) \(\boxed{\phantom{8}}\) \(\times\) DN = 15 \(\times\) 4
\(\therefore\) DN = \(\frac{\boxed{\phantom{60}}}{8}\)
\(\therefore\) DN = \(\boxed{\phantom{7.5}}\)
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