$(5, 3)$, Radius $2$ A1","marks":4,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Rearranges equations for $\\cos \\theta$ and $\\sin \\theta$.","code":"M"},{"id":"p2","marks":1,"description":"Uses the identity $\\cos^2 \\theta + \\sin^2 \\theta = 1$ to eliminate the parameter.","code":"M"},{"id":"p3","marks":1,"description":"Correctly identifies centre $(5, 3)$ and radius $2$.","code":"A"}],"finalAnswers":["Centre (5, 3), Radius 2"],"generatedAt":"2026-05-06T03:46:57.180Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"cartesian_conversion","label":"Cartesian Equation Method","steps":[{"type":"text","order":0,"title":"Isolate trigonometric terms","content":"To convert the parametric equations into a single Cartesian equation, we first rearrange each equation to make $\\cos\\theta$ and $\\sin\\theta$ the subjects.\n\nFor the $x$-equation:\n$$x = 5 + 2\\cos\\theta$$\n$$x - 5 = 2\\cos\\theta$$\n$$\\cos\\theta = \\frac{x - 5}{2}$$\n\nFor the $y$-equation:\n$$y = 3 + 2\\sin\\theta$$\n$$y - 3 = 2\\sin\\theta$$\n$$\\sin\\theta = \\frac{y - 3}{2}$$\n\nThis step earns the first mark M1."},{"type":"text","order":1,"title":"Use Pythagorean identity","content":"Next, we use the fundamental trigonometric identity relating sine and cosine:\n\n$$\\sin^2\\theta + \\cos^2\\theta = 1$$\n\nSubstitute the expressions we found in the previous step into this identity:\n\n$$\\left(\\frac{y - 3}{2}\\right)^2 + \\left(\\frac{x - 5}{2}\\right)^2 = 1$$\n\nThis substitution earns the second mark M1."},{"type":"text","order":2,"title":"Form Cartesian equation","content":"Now, let's simplify the equation to match the standard form of a circle. Square the numerators and denominators:\n\n$$\\frac{(x - 5)^2}{4} + \\frac{(y - 3)^2}{4} = 1$$\n\nMultiply both sides of the equation by $4$ to clear the denominators:\n\n$$(x - 5)^2 + (y - 3)^2 = 4$$\n\nThis standard Cartesian form earns the third mark A1."},{"type":"text","order":3,"title":"Identify centre and radius","content":"The standard Cartesian equation of a circle is given by:\n\n$$(x - h)^2 + (y - k)^2 = r^2$$\n\nwhere $(h, k)$ is the centre and $r$ is the radius.\n\nComparing our equation $(x - 5)^2 + (y - 3)^2 = 4$ to the standard form:\n- The coordinates of the centre are $(h, k) = (5, 3)$.\n- The radius is $r = \\sqrt{4} = 2$.\n\nStating both values correctly earns the final mark A1."}]},{"id":"direct_comparison","label":"Direct Parametric Comparison","steps":[{"type":"text","order":0,"title":"Recall the general form","content":"The general parametric equations for a circle with centre $(h, k)$ and radius $r$ are given by:\n\n$$x = h + r\\cos\\theta$$\n$$y = k + r\\sin\\theta$$\n\nHere, the angle $\\theta$ is the parameter, $r$ is the constant radius, and $(h, k)$ represents the coordinates of the centre."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x = \\htmlData{anchor=h-val}{\\color{@sky}{5}} + \\htmlData{anchor=r1-val}{\\color{@amber}{2}}\\cos\\theta$"},{"latex":"$$y = \\htmlData{anchor=k-val}{\\color{@purple}{3}} + \\htmlData{anchor=r2-val}{\\color{@amber}{2}}\\sin\\theta$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"h-val"},{"color":"purple","style":"text-color","anchor":"k-val"},{"color":"amber","style":"text-color","anchor":"r1-val"},{"color":"amber","style":"text-color","anchor":"r2-val"}],"connections":[{"to":"r2-val","from":"r1-val","color":"amber","label":"radius","style":"dashed"}]},"order":1,"title":"Compare the equations","description":"By comparing our given equations with the general form, we can directly identify the values:\n\n- The horizontal translation $h = 5$\n- The vertical translation $k = 3$\n- The constant coefficient of the trigonometric terms $r = 2$"},{"type":"text","order":2,"title":"State the parameters","content":"From the direct comparison, we get:\n\n* **Centre $(h, k)$:** $(5, 3)$\n* **Radius $r$:** $2$\n\nThis is the final answer! To secure full marks in an exam, we can also

8.2 Using Trigonometric Identities

Question 4

8.2 Using Trigonometric Identities Questions

8.2 Using Trigonometric Identities

Question 4

8.2 Using Trigonometric Identities Questions