.The equation of the required circl...
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.The equation of the required circle is x² + y2 -17x-19y+50 = 0 4. Find the equation of circle passing through intersection points of line ar +by+C =0 with coordinate axes and through origin.

JEE/Engineering Exams
Maths
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( y-cos u s ) we have to find the equation of the given Clecte fou point ( A Rightarrow x=0 Rightarrow a(0) ) thyteeo ( Rightarrow y=-9 / b ) for point ( B Rightarrow y=0 quad ) is ( a x+b(0)+c=0 quad x quad x=-c / a ) We see that onog is fight angled trungle and the Circle is the cion cumenter Civicle of this toriangle. Alvo light argle triangles have a propperty that Thein Cucum center is the midpoin' of theer hypotenuse. So conter of requiried circle ( ddot{1}^{*} Rightarrowleft(frac{-C}{2 a}, frac{-c}{2 b}right) ) Now radius ( r=sqrt{left(frac{c}{2 a}right)^{2}+left(frac{c}{2 b}right)^{2}}=frac{c}{2} sqrt{frac{a^{2}+b^{2}}{a^{2} b^{2}}} ) Wow plaung these values in ( (x-a)^{2}+(y-b)^{2}=x^{2} ) here ( (a, b) ) is center and ( r ) is radis of Cride ( Rightarrowleft(x+frac{c}{2 a}right)^{2}+left(y+frac{c}{2 b}right)^{2}=left(frac{c}{2} sqrt{frac{a^{2}+b^{2}}{a^{2} b^{2}}}right)^{2} ) ( Rightarrow x^{2}+frac{c^{2}}{4 a^{2}}+frac{c x}{a}+y^{2}+left(frac{c}{2 b}right)^{2}+frac{c}{b} y=frac{c^{2}}{4 a^{2}}+frac{c^{2}}{4 b^{2}} Rightarrow x^{2}+y^{2}+frac{c}{a} x+frac{c}{b} y=0 )
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