Question

A CENTER OF THE CIRCLE IS (4a-2,6a+2) AND IS PASSING THE POINT(-6,-2) IF THE DIAMETER OF THE CIRCLE IS 40, THAN FIND THE VALUE OF a .

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Answer 1

$\huge{\underline{\red{\mathcal{ANSWER:}}}}$

${\underline{\bf{Given:}}}$

• $\tt{Center\;of\;circle = (4a - 2,\;6a + 2)}$
• $\tt{Line\;passing\;through\;point = (-6, -2)}$

• $\tt{Diameter\;of\;circle = 40\;units}$
• $\tt{Radius\;of\;circle = 20\;units}$

${\underline{\bf{To\;Find:}}}$

• $\tt{Value\;of\;a.}$

${\underline{\bf{Formula\;used:}}}$

• $\sf{Distance=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\;\;\;\;\;[Distance\;Formula]}$

$\underline{\tt{Now,\;put \;the\;value\;in\;the\;formula,}}$

$\sf{\implies Distance=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}$

$\tt{\implies \sqrt{[4a-2-(-6)]^{2}+[6a+2-(-2)]^{2}}=20}$

$\tt{\implies \sqrt{[4a+4]^{2}+[6a+4]^{2}}=20}$

$\tt{\implies \sqrt{[16a^{2}+16+32a]+[36a^{2}+16+48a]}=20}$

$\tt{\implies \sqrt{16a^{2}+16+32a+36a^{2}+16+48a}=20}$

$\tt{\implies \sqrt{52a^{2}+80a+32}=20}$

$\underline{\tt{Now,\;square\;both\;the\;sides,}}$

$\tt{\implies 52a^{2}+80a+32=400}$

$\tt{\implies 52a^{2}+80a=368}$

$\sf{\implies 52a^{2}+80a-368=0}$

$\underline{\tt{Now, take\;4\;common,\;we\;get}}$

$\tt{\implies 4(13a^{2}+20a-92)=0}$

$\tt{\implies 13a^{2}+20a-92=0}$

$\underline{\tt{Now,\;by\;splitting\;\;middle\;term\;solve\;the\;equation,}}$

$\tt{\implies 13a^{2}+46a-26a-92=0}$

$\tt{\implies a(13a+46)-2(13a+46)=0}$

$\tt{\implies (a-2)(13a+46)=0}$

$\tt{\implies a=2\;or\;-\dfrac{46}{13}}$

$\underline{\tt{Distance\;cannot\;be\;negative.\;Hence,\;a=2.}}$

Answer 2

Step-by-step explanation:

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