Question

A tangent line to a circle is a line that intersects the circle at exactly one point. (It appears to brush the edge of a circle). A point of tangency is the point where a tangent line intersects with a circle. Common external tangents do not intersect the segment that has its endpoints on the centres of the two circles. Common internal tangents intersect the segment that has its end points on the centres of the two circles. The dotted line represent the line segment that has its endpoints on the centres of the sun and the moon.

(i). Identify the tangent lines which partition an area on the earth that experiences a total solar eclipse.

(a) DF and GI

(b) DF and GE

(c) GH and GH

(d) DH and GI.

(ii). From the above drawing, identify the common external tangents between the sun and the moon.

(a) DE and DH

(b) DE and GH

(c) GF and DI

(d) DI and GH.

(iii). From the above drawing, identify the common internal tangents between the sun and the moon.

(a) DE and GE

(b) DH and GH

(c) DE and GH

(d) DH and GE.

(iv). What is the length of DM in the given picture, if EM = y units, HM = 10 units and GM = (y + 10) units ?

(a) 15 units

(b) 20 units

(c) 10.5 units

(d) 10 units.

(v). What is the measure of \(\angle\)DOG in the given picture, if \(\angle\)DMG = x°?

(a) 180°

(b) (180 − x)°

(c) (180 + x)°

(d) 90°.

Answer 1

 (i). It is clear from diagram that tangents DF and GI of sun and moon creates a total solar eclipse on the earth.

Therefore, tangents DF and GI of sun and moon which partition an area on the earth that experiences a total solar eclipse.

Hence, option (a) is correct.

(ii). Since, it is clear by paragraph that the common external tangents do not intersect the segment that has its endpoints on the centres of the two circles.

The segment of the centres of the sun and the moon is OM.

And it is clear by diagram that tangents DE and GH of the sun and the moon does not intersect the segment OM.

Therefore, the common external tangents between the sun and the moon are DE and GH.

Hence, option (b) is correct.

(iii). Since, it is clear by paragraph that the common internal tangents intersect the segment that has its end points on the centres of the two circles.

The segment of the centres of the sun and the moon is OM.

And it is clear by diagram that tangents DH and GE of the sun and the moon intersects the segment OM.

Therefore, the common internal tangents between the sun and the moon are DH and GE.

Hence, option (d) is correct.

(iv). Let the centre of the moon is point o₁.

Let the line DG intersects OM at point A and the line EH intersects Mo₁ at point B.

In triangle ∆DAM and ∆MAG

\(\angle\)DAM = \(\angle\)MAG = 90°

AM = AM (Common line)

 \(\angle\)DMA = \(\angle\)AMG (Angle opposite to the equal line/ common line AM)

Therefore, ∆DAM ≅ ∆MAG. (By SAS congruent rule)

Therefore, DM = GM. (Opposite sides of equal angles)

Thus, DM = GM = y + 10. (∵ GM = (y + 10) units (Given))

Similarly, ∆MBE ≅ ∆MBH.

Therefore, EM = HM. (Opposite sides of equal angles)

⇒ EM = y = 10 units. (∵HM = y units (Given))

Therefore, DM = y + 10 = 10 + 10 = 20 units.

Hence, option (b) is correct.

(v). We know that the angle between tangent and line passes through the centre of a circle is 90°.

Since, DM and GM are tangents and O is centre of the sun, therefore, \(\angle\)ODM = \(\angle\)MGO = 90°.

In quadrilateral OGMD, \(\angle\)ODM + \(\angle\)DMG + \(\angle\)MGO + \(\angle\)DOG = 360°

(Sum of angles of quadrilateral is 360°)

⇒ \(\angle\)DOG + 90° + 90° + x° = 360° (∵ \(\angle\)DMG = x°(given) and \(\angle\)ODM = \(\angle\)MGO = 90°)

⇒ \(\angle\)DOG = 360° − 90° − 90° − x° = 180° − x° = (180 − x)°.

Hence, option (b) is correct.