### Find the coordinates of the mid point of two intersection points of a straight line and a circle

Question Sample Titled 'Find the coordinates of the mid point of two intersection points of a straight line and a circle'

If the striaght line $-{10}{x}+{9}{y}+{7}={0}$ and the circle ${x}^{{{2}}}+{y}^{{{2}}}-\dfrac{{335}}{{49}}{x}-\dfrac{{115}}{{49}}{y}-\dfrac{{236}}{{7}}={0}$ intersect at ${A}{\left({x}_{{1}},{y}_{{1}}\right)}$ and ${B}{\left({x}_{{2}},{y}_{{2}}\right)}$, then the ${y}$-coordinate of the mid-point of ${A}{B}=$

A
${2}$
B
$\dfrac{{5}}{{2}}$
C
${1}$
D
$\dfrac{{5}}{{4}}$

 ${x}$ $=\dfrac{{9}}{{10}}{y}+\dfrac{{7}}{{10}}$ $\ldots{\left({1}\right)}$ ${x}^{{{2}}}+{y}^{{2}}-\dfrac{{335}}{{49}}{x}-\dfrac{{115}}{{49}}{y}-\dfrac{{236}}{{7}}$ $={0}$ $\ldots{\left({2}\right)}$ Substitute ${\left({1}\right)}$ into ${\left({2}\right)}:$ ${\left(\dfrac{{9}}{{10}}{y}+\dfrac{{7}}{{10}}\right)}^{{{2}}}+{y}^{{2}}-\dfrac{{335}}{{49}}{\left(\dfrac{{9}}{{10}}{y}+\dfrac{{7}}{{10}}\right)}-\dfrac{{115}}{{49}}{y}-\dfrac{{236}}{{7}}$ $={0}$ $\dfrac{{81}}{{100}}{y}^{{2}}+\dfrac{{63}}{{50}}{y}+\dfrac{{49}}{{100}}+{y}^{{2}}-\dfrac{{603}}{{98}}{y}-\dfrac{{67}}{{14}}-\dfrac{{115}}{{49}}{y}-\dfrac{{236}}{{7}}$ $={0}$ $\dfrac{{181}}{{100}}{y}^{{2}}-\dfrac{{181}}{{25}}{y}-\dfrac{{3801}}{{100}}$ $={0}$ ${\left({y}-{7}\right)}{\left({y}+{3}\right)}$ $={0}$ ${y}$ $={7}$ or ${y}=-{3}$ ∴   The ${y}$-coordinates of ${A}$ and ${B}$ are ${7}$ and $-{3}$ respectively. The ${y}$-coordinate of the mid-point of AB $=\dfrac{{{7}-{3}}}{{2}}$ $={2}$

 Note that the line joining centre to mid-point of chord is perpendicular to chord. Centre of circle $={\left(\dfrac{{-\dfrac{{335}}{{49}}}}{{-{{2}}}},\dfrac{{-\dfrac{{115}}{{49}}}}{{-{{2}}}}\right)}={\left(\dfrac{{335}}{{98}},\dfrac{{115}}{{98}}\right)}$ Denote ${L}$ be the straight line which passes through the mid-point of ${A}{B}$ and the centre of the circle. Slope of ${A}{B}=\dfrac{{10}}{{9}}$ ∵  ${A}{B}\bot{L}$ , slope of ${L}=\dfrac{{-{1}}}{{\dfrac{{10}}{{9}}}}=-\dfrac{{9}}{{10}}$ The equation of ${L}:$ ${y}-\dfrac{{115}}{{98}}=-\dfrac{{9}}{{10}}{\left({x}-\dfrac{{335}}{{98}}\right)}$ Point slope form ${L}:$ ${y}$ $=-\dfrac{{9}}{{10}}{x}+\dfrac{{17}}{{4}}$ $\ldots{\left({1}\right)}$ ${A}{B}:$ $-{10}{x}+{9}{y}+{7}$ $={0}$ $\ldots{\left({2}\right)}$ Substitute ${\left({1}\right)}$ into ${\left({2}\right)}:$ $-{10}{x}+{9}{\left(-\dfrac{{9}}{{10}}{x}+\dfrac{{17}}{{4}}\right)}+{7}$ $={0}$ $-{10}{x}-\dfrac{{81}}{{10}}{x}+\dfrac{{153}}{{4}}+{7}$ $={0}$ $-\dfrac{{181}}{{10}}{x}+\dfrac{{181}}{{4}}$ $={0}$ ${x}$ $=\dfrac{{5}}{{2}}$ Substitute ${x}$ $=\dfrac{{5}}{{2}}$ into ${\left({1}\right)}:$ ${y}$ $=-\dfrac{{9}}{{10}}{\left(\dfrac{{5}}{{2}}\right)}+\dfrac{{17}}{{4}}$  $={2}$ ∴   The ${y}$-coordinate of the mid-point of ${A}{B}$ is ${2}.$

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