The circle x^2 +y^2 −2x−4y−20=0 is inscribed in a square. One vertex of the square is (−4, 7). Find the coordinates of the other vertices.


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Question Number 174185 by nadovic last updated on 26/Jul/22

$The circle x^{2} + y^{2} - 2 x - 4 y - 20 = 0 is$ $inscribed in a square . One vertex$ $of the square is (- 4, 7) . Find the$ $coordinates of the other vertices .$
	Answered by Cesar1994 last updated on 26/Jul/22

	$x^{2} + y^{2} - 2 x - 4 y - 20 = 0 \Leftrightarrow {(x - 1)}^{2} + {(y - 2)}^{2} = 20 + 1 + 4$ $\Leftrightarrow {(x - 1)}^{2} + {(y - 2)}^{2} = 25$ $t h e {c i r c l e}^{'} s c e n t e r i s (1, 2)$ $(\frac{- 3 - 7}{6 - (- 4)}) (\frac{- 3 - 7}{- 4 - 6}) = \frac{- 10}{10} (\frac{- 10}{- 10}) = - 1$ $t h e d i a m e t r i c a l l y o p o s i t e v e r t e x t o A (- 4, 7) i s B (1 + 5, 2 - 5) = B (6, - 3)$ $t h e o t h e r s v e r t i c e s a r e C (1 + 5, 2 + 5) = C (6, 7) a n d D (1 - 5, 2 - 5) = D (- 4, - 3)$ $e v a l u a t i n g e a c h p o i n t i n (1) i t i s v e r i f i e d t h a t t h e y a r e o n t h e c i r c u m f e r e n c e$ $t h e n p r o v e t h a t t h e d i a g o n a l s A B a n d C D a r e p e r p e n d i c u l a r s o n e t o t h e o t h e r$ $(\frac{- 3 - 7}{6 - (- 4)}) (\frac{- 3 - 7}{- 4 - 6}) = \frac{- 10}{10} (\frac{- 10}{- 10}) = - 1$ $∴ t h e A B C D i s r h o m b u s$ $t h e n s i d e A C i s h o r i z o n t a l a n d s i d e B C i s v e r t i c a l s i d e, s o t h e y a r e p e r p e n d i c u l a r$ $∴ A B C D i s a s q u a r e ◼$
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