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Q39E

Expert-verifiedFound in: Page 7

Book edition
5th

Author(s)
Otto Bretscher

Pages
442 pages

ISBN
9780321796974

**Find the circle that runs through the points (5,5),(4,6), ****and (6,2)****. Write your equation in the form ${\mathit{a}}{\mathbf{+}}{\mathit{b}}{\mathit{x}}{\mathbf{+}}{\mathit{c}}{\mathit{y}}{\mathbf{+}}{{\mathit{x}}}^{{\mathbf{2}}}{\mathbf{+}}{{\mathit{y}}}^{{\mathbf{2}}}{\mathbf{=}}{\mathbf{0}}$****. Find the centre and radius of this circle.**

The required equation is ${x}^{2}+{y}^{2}-2x-4y=20$, which is the circle with centre (1,2) and radius 5 that runs through the points (5,5),(4,6) and (6,2).

**The centre and the radius of the circle of form ${\mathbf{(}}{\mathit{x}}{\mathbf{-}}{\mathit{a}}{\mathbf{)}}^{\mathbf{2}}{\mathbf{+}}{\mathbf{(}}{\mathit{y}}{\mathbf{-}}{\mathit{b}}{\mathbf{)}}^{\mathbf{2}}{\mathbf{=}}{{\mathit{r}}}^{{\mathbf{2}}}$**** is ${\left(a,b\right)}$and r respectively.**

Substitute the given points in the equation $a+bx+cy+{x}^{2}+{y}^{2}=0$

${5}^{2}+{5}^{2}+a+\left(5\right)b+\left(5\right)c=0\phantom{\rule{0ex}{0ex}}a+5b+5c=-50$

${4}^{2}+{6}^{2}+a+\left(4\right)b+\left(6\right)c=0\phantom{\rule{0ex}{0ex}}a+4b+6c=-52$

${6}^{2}+{2}^{2}+a+\left(6\right)b+\left(2\right)c=0\phantom{\rule{0ex}{0ex}}a+6b+2c=-40$

Consider the simplified equations.

$-50=a+5b+5c......\left(1\right)\phantom{\rule{0ex}{0ex}}-52=a+4b+6c......\left(2\right)\phantom{\rule{0ex}{0ex}}-40=a+6b+2c......\left(3\right)$

Arrange the equations (1), (2) and (3) into matrix form:

$\left(\begin{array}{ccc}1& 5& 5\\ 1& 4& 6\\ (1& 6& 2\end{array}\right)\left(\begin{array}{c}a\\ b\\ c\end{array}\right)=\left(\begin{array}{c}-50\\ -52\\ -40\end{array}\right)$

Upon solving the values of $a,b$ and are obtained as $a=-20,b=-2,c=-4$

Substitute these values in the given equation of the circle.

${x}^{2}+{y}^{2}+\left(-20\right)+\left(-2\right)x+\left(-4\right)y=0\phantom{\rule{0ex}{0ex}}{x}^{2}+{y}^{2}-20-2x-4y=0$

Rearrange the resulting equation into standard form of circle as:

${x}^{2}+{y}^{2}-20-2x-4y=0\phantom{\rule{0ex}{0ex}}{x}^{2}+{y}^{2}-2x-4y=20\phantom{\rule{0ex}{0ex}}({x}^{2}+1-2x)+({y}^{2}+4-4y)=20+1+4\phantom{\rule{0ex}{0ex}}(x-1{)}^{2}+(y-2{)}^{2}=25$

Compare the above equation with the standard equation of the circle.

${\left(x-a\right)}^{2}+{\left(y-b\right)}^{2}={r}^{2}\iff {\left(x-1\right)}^{2}+{\left(y-2\right)}^{2}={5}^{2}\phantom{\rule{0ex}{0ex}}\left(a,b\right)=\left(1,2\right)\phantom{\rule{0ex}{0ex}}r=5$

The required equation is ${x}^{2}+{y}^{2}-2x-4y=20$, which is the circle with centre (1,2) and radius 5 that runs through the points (5,5),(4,6) and (6,2).

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