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Question Number 61003 by Tawa1 last updated on 28/May/19

	Answered by tanmay last updated on 28/May/19

	$P Q = d i a m e t e r = \sqrt{{(4 - 0)}^{2} + {(0 - 2)}^{2}} = 2 \sqrt{5}$ $r a d i u s = \sqrt{5}$ $c e n t r e = (\frac{0 + 4}{2}, \frac{2 + 0}{2}) \to (2, 1)$ $e q n c i r c l e {(x - 2)}^{2} + {(y - 1)}^{2} = {(\sqrt{5})}^{2}$ $x^{2} + y^{2} - 4 x - 2 y = 0$
	Commented by Tawa1 last updated on 28/May/19

	$God bless you sir$
	Answered by tanmay last updated on 28/May/19

	$i i) e q n c i r c l e x^{2} + y^{2} - 4 x - 2 y = 0$ $e q n o f t a n g e n t a t Q (4, 0)$ $x (4) + y (0) - 2 (x + 4) - (y + 0) = 0$ $4 x - 2 x - 8 - y = 0$ $2 x - y = 8$ $f o r m u l a t o f i n d t a n g e n t a t p o i n t (x_{1}, y_{1})$ $w h e n (x_{1}, y_{1}) l i e s o n c i r c l e$ $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ $t r i c k s . . .$ $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ $x \times x + y \times y + g (x + x) + f (y + y) + c = 0$ $n o w$ ${x x}_{1} + {y y}_{1} + g (x + x_{1}) + f (y + y_{1}) + c = 0 i s t a n g e n t$ $t h e e q n o f d i a m e t e r ∥ t o 2 x - y = 8$ $i s 2 x - y = k p a s s i n g t h r o u g h c d n t r e o f c i r c l e$ $s o 2 \times 2 - 1 = k \to k = 3$ $s o t h e r e a u i r e d d i a m e t e r ∥ t o 2 x - y = 8$ $i s 2 x - y = 3$
	Commented by Tawa1 last updated on 28/May/19

	$God bless you sir$
	Answered by tanmay last updated on 28/May/19

	$c i r c l e x^{2} + y^{2} - 4 x - 2 y = 0$ $c e n t r e = (2, 1) r a d i u z = \sqrt{5}$ $d i s t a n c e f r o m (2, 1) t o 2 y + x + 1 = 0$ $i s ∣ \frac{2 \times 1 + 2 + 1}{\sqrt{2^{2} + 1^{2}}} ∣$ $= \sqrt{5}$ $= t h e r a d i u s o f c i r c l e$ $s o 2 y + x + 1 = 0 i s t a n g e n t t o c i r c l e x^{2} + y^{2} - 4 x - 2 y = 0$
	Commented by Tawa1 last updated on 28/May/19

	$God bless you sir$
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