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"variablesTest": {"condition": "", "maxRuns": 100}, "statement": "\n    <p>Find the points where the circles:</p>\n  <ul>\n  <li>$\\simplify[std]{x^2+y^2 -{2*k2}x -{2*l2}y+{2*k2*a+2*l2*b-a^2-b^2}}=0$\n  </li>\n  </ul>\n  <p>&nbsp;</p>\n  <ul>\n  <li>and $\\simplify[std]{x^2+y^2 -{2*k}x -{2*l}y+{2*k*a+2*l*b-a^2-b^2}}=0$</li>\n  </ul>\n  <p>&nbsp;</p>\n  meet.\n  <p>&nbsp;</p>\n  \n  <p>Enter the coordinates for the two points, starting with the point with the smaller $x$-coordinate.</p>\n  <p>Note that, for example, $-3$ is smaller than $-1$. Thus if the points of intersection were $(-1, 4)$ and $(-3,6)$ you would enter $(-3,6)$ for the first point and $(-1, 4)$ for the second.</p>\n  <p>Input the coordinates as integers.</p>\n  \n  \n    ", "tags": ["checked2015", "equation of a circle", "equation of a straight line", "Equation of a straight line", "intersection of two circles", "quadratic equation"], "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "<p>Find the points of intersection of two circles.</p>"}, "extensions": [], "advice": "\n    <p>Subtract the equation for the first circle from the equation for the second. You get $\\simplify[std]{{2*k2-2*k}x+ {2*l2-2*l}y+{2*k*a+2*l*b-2*k2*a-2*l2*b}}=0$.</p>\n    <p>This can be rearranged as: $y = \\simplify[std]{({b-d}/{a-c})x+{b*c-a*d}/{c-a}}$</p>\n    <p>Substitute for $y = \\simplify[std]{({b-d}/{a-c})x+{b*c-a*d}/{c-a}}$ in the equation of the circle.</p>\n  <p>We get $\\simplify[std]{x^2+(({b-d}/{a-c})x+{b*c-a*d}/{c-a})^2 -{2*k}x -{2*l}(({b-d}/{a-c})x+{b*c-a*d}/{c-a})+{2*k*a+2*l*b-a^2-b^2}}=0$.</p>\n  <p>Collecting terms we get $\\simplify[std]{{1+ (b-d)^2/(a-c)^2} x^2+{(-2)*(b-d)(b*c-a*d)/(a-c)^2-2*k  -(2*l)*(b-d)/(a-c)}x+{(b*c-a*d)^2/(c-a)^2 -2*l*(b*c-a*d)/(c-a)+2*k*a+2*l*b-a^2-b^2}}=0$.</p>\n  <p>Solving this quadratic equation for $x$ gives solutions $x=\\var{a}$ and $x=\\var{c}$.</p>\n  <p>Substitute these values of $x$ into the equation $\\simplify[std]{({b-d}/{a-c})x+{b*c-a*d}/{c-a}}$ to obtain the $y$-coordinates: $y=\\var{b}$ (when $x=\\var{a}$) and $y=\\var{d}$ (when $x=\\var{c}$).</p>\n  <p>Enter the coordinate pairs in the order $(\\var{m},\\var{n})$, $(\\var{p}, \\var{q})$. </p>\n    ", "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}]}]}], "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}]}