{"@context": "https://schema.org/", "@type": "Quiz", "about": {"@type": "Thing", "name": "Geometrical Progression" }, "educationalAlignment": [{"@type": "AlignmentObject","alignmentType": "educationalSubject","targetName": "Mathematics"}], "hasPart": [{"@context": "https://schema.org/","@type": "Question","eduQuestionType": "Flashcard", "text": "The abscissa of 2 points A and B are the roots of the equation x^2+2x-a^2=0 and the ordinate are the roots of the equation y^2+4y-b^2=0. Find the equation of the circle with AB as a diameter. - ReadAxis", "acceptedAnswer": {"@type": "Answer", "text": "Let roots of the equation x2+2x–a2=0 are x1andx2and roots of the equation y2+4y–b2=0 are y1andy2Therefore x1+x2=−21andx1x2=−a21Similarly y1+y2=−41andy1y2=−b21Hence abscissa and ordinate of A are x1andy1 respectively.Similarly for B are x2andy2.we know the diametric form of circle is (y–y1)(y–y2)=−(x–x1)(x–x2)⇒x2–(x1+x2)x+x1x2+y2–(y1+y2)y+y1y2=0⇒x2−(−2)x+(−a2)+y2–(−4)y+(−b2)=0Hence Equation required = x2+y2+2x+4y–(a2+b2)=0Here is the Second and Easy Method:If roots of two quadratic equations say ap2+bp+c=0 and dq2+eq+f=0, represent endpoints(one equation their abscissa and other ordinates) of a diameter of the circle, then the equation of the circle will be the summation of both the equations after making their second-degree term coefficient = 1.That is C : p2+bap+ca+q2+edq+fd=0" }}]} {"@context": "http://schema.org","@type": "QAPage", "name": "The abscissa of 2 points A and B are the roots of the equation x^2+2x-a^2=0 and the ordinate are the roots of the equation y^2+4y-b^2=0. Find the equation of the circle with AB as a diameter. - ReadAxis", "description": "Let roots of the equation x2+2x–a2=0 are x1andx2and roots of the equation y2+4y–b2=0 are y1andy2Therefore x1+x2=−21andx1x2=−a21Similarly y1+y2=−41andy1y2=−b21Hence abscissa and ordinate of A are x1andy1 respectively.Similarly for B are x2andy2.we know the diametric form of circle is (y–y1)(y–y2)=−(x–x1)(x–x2)⇒x2–(x1+x2)x+x1x2+y2–(y1+y2)y+y1y2=0⇒x2−(−2)x+(−a2)+y2–(−4)y+(−b2)=0Hence Equation required = x2+y2+2x+4y–(a2+b2)=0Here is the Second and Easy Method:If roots of two quadratic equations say ", "mainEntity": {"@type": "Question", "@id": "https://readaxis.com/question-answer/the-abscissa-of-2-points-a-and-b-are-the-roots-of-the-equation-x22x-a20-and-the-ordinate-are-the-roots-of-the-equation-y24y-b20-find-the-equation-of-the-circle-with-ab-as-a-diameter/", "name": "The abscissa of 2 points A and B are the roots of the equation x^2+2x-a^2=0 and the ordinate are the roots of the equation y^2+4y-b^2=0. Find the equation of the circle with AB as a diameter. - ReadAxis", "text": "The abscissa of 2 points A and B are the roots of the equation x^2+2x-a^2=0 and the ordinate are the roots of the equation y^2+4y-b^2=0. Find the equation of the circle with AB as a diameter. - ReadAxis", "dateCreated": "2023-01-20T07:58:18.562Z", "answerCount": "1", "author": { "@type": "Person", "name": "ReadAxis", "url": "https://www.readaxis.com/" }, "acceptedAnswer": { "@type": "Answer", "upvoteCount": "5", "text": "Let roots of the equation x2+2x–a2=0 are x1andx2and roots of the equation y2+4y–b2=0 are y1andy2Therefore x1+x2=−21andx1x2=−a21Similarly y1+y2=−41andy1y2=−b21Hence abscissa and ordinate of A are x1andy1 respectively.Similarly for B are x2andy2.we know the diametric form of circle is (y–y1)(y–y2)=−(x–x1)(x–x2)⇒x2–(x1+x2)x+x1x2+y2–(y1+y2)y+y1y2=0⇒x2−(−2)x+(−a2)+y2–(−4)y+(−b2)=0Hence Equation required = x2+y2+2x+4y–(a2+b2)=0Here is the Second and Easy Method:If roots of two quadratic equations say ap2+bp+c=0 and dq2+eq+f=0, represent endpoints(one equation their abscissa and other ordinates) of a diameter of the circle, then the equation of the circle will be the summation of both the equations after making their second-degree term coefficient = 1.That is C : p2+bap+ca+q2+edq+fd=0", "url": "https://readaxis.com/question-answer/the-abscissa-of-2-points-a-and-b-are-the-roots-of-the-equation-x22x-a20-and-the-ordinate-are-the-roots-of-the-equation-y24y-b20-find-the-equation-of-the-circle-with-ab-as-a-diameter/", "dateCreated": "2023-01-20T07:58:18.562Z", "author": { "@type": "Person", "name": "Readaxis Admin" } }, "suggestedAnswer": [] } }

Question

The abscissa of 2 points A and B are the roots of the equation \(x^2 + 2x – a^2 = 0\) and the ordinate are the roots of the equation \(y^2 + 4y – b^2 = 0\). Find the equation of the circle with AB as a diameter.

Solution:

Let roots of the equation \(x^2 + 2x – a^2 = 0\) are \(x_1 \; and\; x_2\)
and roots of the equation \(y^2 + 4y – b^2 = 0\) are \(y_1 \; and \; y_2\)

Therefore \(x_1 + x_2 = \frac{-2}{1} \;and \;x_1x_2 = \frac{-a^2}{1}\)
Similarly \(y_1 + y_2 = \frac{-4}{1} \;and\; y_1y_2 = \frac{-b^2}{1}\)

Hence abscissa and ordinate of A are \(x_1\; and \; y_1\) respectively.
Similarly for B are \(x_2\;and\;y_2\).

we know the diametric form of circle is \((y – y_1)(y – y_2) = -(x – x_1)(x – x_2)\)
\(\Rightarrow x^2 – (x_1 + x_2)x + x_1x_2 + y^2 – (y_1 + y_2)y + y_1y_2 = 0\)
\(\Rightarrow x^2 -(-2)x + (-a^2) + y^2 – (-4)y + (-b^2) = 0 \)

Hence Equation required = \(x^2 + y^2 + 2x + 4y – (a^2 + b^2) = 0 \)

Here is the Second and Easy Method:

If roots of two quadratic equations say \(ap^2 + bp + c = 0\) and \(dq^2 + eq + f = 0\), represent endpoints(one equation their abscissa and other ordinates) of a diameter of the circle, then the equation of the circle will be the summation of both the equations after making their second-degree term coefficient = 1.

That is C : \(p^2 + \frac{b}{a}p + \frac{c}{a} + q^2 + \frac{e}{d}q + \frac{f}{d} = 0\)