Binomial Theorem

Question
CBSEENMA11015080

Find the centre and radii of the following circle:
open parentheses straight x plus 3 over 2 close parentheses squared plus open parentheses straight y minus 5 over 2 close parentheses squared equals 289 over 4

Solution

The given equation of the circle is: open parentheses straight x plus 3 over 2 close parentheses squared plus open parentheses straight y minus 5 over 2 close parentheses squared equals 289 over 4
Comparing it with left parenthesis straight x minus straight h right parenthesis squared plus left parenthesis straight y minus straight k right parenthesis squared equals straight r squared comma space we space get
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493)
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#4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array)
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
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  ∴ the centre of circle is left parenthesis straight h comma space straight k right parenthesis space left right arrow space open parentheses negative 3 over 2 comma space 5 over 2 close parentheses space and space the space radius space is space 17 over 2
The radius of circle is straight r equals 17 over 2

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