As per the inquiry,
Two circles with focuses O and O’ of radii 3 cm and 4 cm, individually meet at two focuses P and Q, to such an extent that OP and O’P are digressions to the two circles and PQ is a typical harmony.
To Find: Length of normal harmony PQ
∠OPO’ = 90° [Tangent at a point on the circle is opposite to the sweep through mark of contact]
So OPO is a right-calculated triangle at P
Utilizing Pythagoras in △ OPO’, we have
![\[\left( OO' \right)2=\text{ }\left( O'P \right)2+\text{ }\left( OP \right)2\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c17f7eb418d487c6c0b1188dcd6e65e3_l3.png "Rendered by QuickLaTeX.com")
![\[\left( OO' \right)2\text{ }=\text{ }\left( 4 \right)2\text{ }+\text{ }\left( 3 \right)2\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d936d8fef1b560fb3a499d028838f080_l3.png "Rendered by QuickLaTeX.com")
![\[\left( OO' \right)2\text{ }=\text{ }25\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-32030d19f07a1cd76c66645d92c82b6a_l3.png "Rendered by QuickLaTeX.com")
OO’ = 5 cm
Let ON = x cm and NO’ = 5 – x cm
In right calculated triangle ONP
![\[\left( ON \right)2+\text{ }\left( PN \right)2=\text{ }\left( OP \right)2\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-07a29178da0381dcb34e0db6892e71a9_l3.png "Rendered by QuickLaTeX.com")
![\[x2\text{ }+\text{ }\left( PN \right)2\text{ }=\text{ }\left( 3 \right)2\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a34eeddd3e44c4289cf8e4bd4f1563fb_l3.png "Rendered by QuickLaTeX.com")
![\[\left( PN \right)2=\text{ }9\text{ }\text{ }x2\text{ }\left[ 1 \right]\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c1a4266ad674b1c4fd9b7aa7f76fccc5_l3.png "Rendered by QuickLaTeX.com")
In right calculated triangle O’NP
![\[\left( O'N \right)2\text{ }+\text{ }\left( PN \right)2=\text{ }\left( O'P \right)2\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-583e3f55ed70e1a532eabaffca792194_l3.png "Rendered by QuickLaTeX.com")
![\[\left( 5\text{ }\text{ }x \right)2\text{ }+\text{ }\left( PN \right)2\text{ }=\text{ }\left( 4 \right)2\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3b68a7acb4bb351d2deb572316e9638e_l3.png "Rendered by QuickLaTeX.com")
![\[25\text{ }\text{ }10x\text{ }+\text{ }x2\text{ }+\text{ }\left( PN \right)2\text{ }=\text{ }16\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fea28d3a3c8363def7fa03ec597eeba3_l3.png "Rendered by QuickLaTeX.com")
![\[\left( PN \right)2\text{ }=\text{ }-\text{ }x2+\text{ }10x\text{ }\text{ }9\left[ 2 \right]\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9e79aa1019c6edf45a5c987513195b4d_l3.png "Rendered by QuickLaTeX.com")
From [1] and [2]
![\[9\text{ }\text{ }x2\text{ }=\text{ }-\text{ }x2\text{ }+\text{ }10x\text{ }\text{ }9\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-62067c4fb6f9690aa7f0cd34b1361805_l3.png "Rendered by QuickLaTeX.com")
10x = 18
x = 1.8
From (1) we have
![\[\left( PN \right)2\text{ }=\text{ }9\text{ }\text{ }\left( 1.8 \right)2\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-eeb728de19bd4159054365047081c0f7_l3.png "Rendered by QuickLaTeX.com")
![\[=9\text{ }\text{ }3.24\text{ }=\text{ }5.76\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6cf89b4ba86df50d789bbb39ca746e85_l3.png "Rendered by QuickLaTeX.com")
PN = 2.4 cm
![\[PQ\text{ }=\text{ }2PN\text{ }=\text{ }2\left( 2.4 \right)\text{ }=\text{ }4.8\text{ }cm\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-48556d6216d5899e02edcaa3bc9cf44f_l3.png "Rendered by QuickLaTeX.com")
