Math Question and Answers


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 200356 by sonukgindia last updated on 17/Nov/23

Commented by mr W last updated on 17/Nov/23

$i g u e s s y o u m e a n t t h a t t h e p l a n e$ $x + 8 y - 4 z + k = 0 s h o u l d t o u c h t h e$ $s p h e r e x^{2} + y^{2} + z^{2} + 2 y - 3 = 0 .$
	Answered by mr W last updated on 17/Nov/23

	$x^{2} + {(y + 1)}^{2} + z^{2} = 2^{2}$ $c e n t e r o f s p h e r e a t (0, - 1, 0)$ $r a d i u s o f s p h e r e i s 2 .$ $d i s t a n c e f r o m (0, - 1, 0) t o t h e p l a n e$ $s h o u l d b e 2 .$ $\frac{1 \times 0 + 8 \times (- 1) - 4 \times 0 + k}{\sqrt{1^{2} + 8^{2} + {(- 4)}^{2}}} = \pm 2$ $k - 8 = \pm 2 \times 9$ $\Rightarrow k = 8 \pm 18 = - 10 o r 26$ $Σ k = - 10 + 26 = 16 ✓$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum