### equation of st. lines and circle

Question Sample Titled 'equation of st. lines and circle'

The equation of the straight line ${L}_{{1}}$ is ${x}+{4}{y}={0}$. ${L}_{{2}}$ is another straight line passing through point ${A}{\left(-{9},{3}\right)}$ and perpendicular to ${L}_{{1}}$.

 (a) Find the equation of ${L}_{{2}}$. (2 marks) (b) Suppose that ${G}$ is a point lying on ${L}_{{2}}$. Denote the ${y}$-coordinate of ${G}$ by ${k}$, where ${k}\ne{3}$. Let ${C}$ be the circle passing through ${A}$ with centre ${G}$. (4 marks) Prove that the equation of ${C}$ is ${2}{x}^{{2}}+{2}{y}^{{2}}+{\left({39}-{k}\right)}{x}-{4}{k}{y}+{3}{k}+{171}={0}$. (c) The coordinates of the point ${B}$ are (1,1). Using (b), or therwise, find the radius of the circle which passes through ${A}$ and ${B}$ with centre ${G}$. (3 marks)

 (a) Slope of ${L}_{{1}}$ $=-\dfrac{{1}}{{4}}$ ∵  ${L}_{{1}}\bot{L}_{{2}}.$ ∴   Slope of ${L}_{{2}}=-{\left(-\dfrac{{1}}{{4}}\right)}^{{-{{1}}}}$ 1M  $={4}$ Equation of ${L}_{{2}}:$ $\dfrac{{{y}-{3}}}{{{x}-{\left(-{9}\right)}}}$ $={4}$ ${y}-{3}$ $={4}{x}+{36}$ ${4}{x}-{y}+{39}$ $={0}$ 1A  (b) Sub. ${y}={k}$ into the equation of ${L}_{{2}}$, we have ${x}$ $=\dfrac{{{k}-{39}}}{{4}}$ 1M Thus, ${G}={\left(\dfrac{{{k}-{39}}}{{4}},{k}\right)}$ Radius of ${C}$ $=\sqrt{{{\left(\dfrac{{{k}-{39}}}{{4}}-{\left(-{9}\right)}\right)}^{{2}}+{\left({k}-{3}\right)}^{{2}}}}$ 1M  $=\sqrt{{{\left(\dfrac{{{k}-{3}}}{{4}}\right)}^{{2}}+{\left(\dfrac{{{4}{\left({k}-{3}\right)}}}{{4}}\right)}^{{2}}}}$  $=\dfrac{{1}}{{4}}\sqrt{{{\left({k}-{3}\right)}^{{2}}+{16}{\left({k}-{3}\right)}^{{2}}}}$  $=\pm\dfrac{{\sqrt{{{17}}}}}{{4}}{\left({k}-{3}\right)}$

 Equation of ${C}:$ ${\left({x}-\dfrac{{{k}-{39}}}{{4}}\right)}^{{2}}+{\left({y}-{k}\right)}^{{2}}$ $={\left(\pm\dfrac{{\sqrt{{{17}}}}}{{4}}{\left({k}-{3}\right)}\right)}^{{2}}$ 1M ${x}^{{2}}-{2}{x}{\left(\dfrac{{{k}-{39}}}{{4}}\right)}+{\left(\dfrac{{{k}-{39}}}{{4}}\right)}^{{2}}+{y}^{{2}}-{2}{k}{y}+{k}^{{2}}$ $=\dfrac{{17}}{{16}}{\left({k}-{3}\right)}^{{2}}$ 1M ${x}^{{2}}+{y}^{{2}}+{\left(\dfrac{{{39}-{k}}}{{2}}\right)}{x}+\dfrac{{{k}^{{2}}-{78}{k}+{39}^{{2}}}}{{16}}-{2}{k}{y}+{k}^{{2}}-\dfrac{{17}}{{16}}{\left({k}-{3}\right)}^{{2}}$ $={0}$ ${x}^{{2}}+{y}^{{2}}+{\left(\dfrac{{{39}-{k}}}{{2}}\right)}{x}-{2}{k}{y}+\dfrac{{{17}{k}^{{2}}-{78}{k}+{39}^{{2}}}}{{16}}-\dfrac{{{17}{k}^{{2}}-{102}{k}+{153}}}{{16}}$ $={0}$ ${x}^{{2}}+{y}^{{2}}+{\left(\dfrac{{{39}-{k}}}{{2}}\right)}{x}-{2}{k}{y}+\dfrac{{{24}{k}+{1368}}}{{16}}$ $={0}$ ${x}^{{2}}+{y}^{{2}}+{\left(\dfrac{{{39}-{k}}}{{2}}\right)}{x}-{2}{k}{y}+\dfrac{{{3}{k}+{171}}}{{2}}$ $={0}$ ${2}{x}^{{2}}+{2}{y}^{{2}}+{\left({39}-{k}\right)}{x}-{4}{k}{y}+{3}{k}+{171}$ $={0}$

 (c) Sub. ${\left({1},{1}\right)}$ into the equation of ${C}$, ${2}+{2}+{39}-{k}-{4}{k}+{3}{k}+{171}$ $={0}$ ${214}$ $={2}{k}$ ${k}$ $={107}$ 1A ∵   Radius of ${C}$ $=\pm\dfrac{{\sqrt{{{17}}}}}{{4}}{\left({k}-{3}\right)}$ Thus, Radius of ${C}$ $=\dfrac{{\sqrt{{{17}}}}}{{4}}{\left({107}-{3}\right)}$  $={26}\sqrt{{{17}}}$ $\text{units}$

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