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प्रश्न
Two circles with centres O and P intersect each other at A and B as shown in following fig. Two straight lines MAN and RBQ are drawn parallel to OP.
Prove that (i) MN = 20 P (ii) MN= RQ.
उत्तर
Given: Two cirdes with centres 0 and P, and MN II OP || RQ
To prove: (i) MN = 20P (ii) MN= RQ.
Construction: OX ⊥ MN, PY ⊥ MN, OW ⊥ RZ, PZ ⊥ RQ
Proof: Since each angle of the quadrilateral XYZW is a right angle, sc XYZW is a rectangle.
Also, XYPO is a rectangle. ...(1)
Now, perpendicular drawn from the centre to the chord bisects the chord.
Therefore, MA = 2 XA and AN = 2 AY ...(2)
Now, MN = MA + AN = 2 XA + 2 AY [from (2)]
⇒ MN = 2(XA + AY) = 2 XY
⇒ MN = 2 OP [As XYPO is a rectangle, XY = OP] ... (3)
This proves part (i).
By similar arguments, we have RQ = 2 OP ...(4)
Using (3) and ( 4), we get
MN= RQ.
This proves part (ii).
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Two circles with centres O and O' touch each other at point L. Prove that, a tangent through L bisects the common tangent AB of the two circles at point M.
Given: AB is a common tangent of the two circles that touch each other at point L. ML is a tangent through point L.
To prove: M is a mid-point of the tangent AB or MA = MB.
Proof: From the figure,
M is an external point that draws two tangents, MA and ML to the circle with the centre O.
So, `square` = `square` ......(i)
Similarly, M draws two tangents ML and MB to the circle with the centre O'.
So, `square` = `square` ......(ii)
From the equations (i) and (ii),
`square` = `square`
Hence, the tangent at the point L, bisects the common tangent, AB of the two circles at point M.
