Circles
Tangent to Circle
Grade 11
Question:
<p>Let point <span>P</span>(<span>x</span><sub>1</sub>, <span>y</span><sub>1</sub>) be any point on the circle <span>(x</span><sub>1</sub> - 3)<sup>2</sup> + (<span>y</span><sub>1</sub> + 2)<sup>2</sup> = 5<span>r</span><sup>2</sup>. Find the area between two circles if the length of tangent drawn from point <span>P</span>(<span>x</span><sub>1</sub>, <span>y</span><sub>1</sub>) to the circle <span>(x</span> - 3)<sup>2</sup> + (<span>y</span> + 2)<sup>2</sup> = <span>r</span><sup>2</sup> is such that the area is <span>k</span>π.</p>
Step-by-Step Solution
Key Concept: Use the tangent length formula and the constraint that P lies on both circles to find radius r, then calculate the area difference.
<p><strong>Step 1:</strong> Since point P(<span>x</span><sub>1</sub>, <span>y</span><sub>1</sub>) lies on the circle, it satisfies: <span>(x</span><sub>1</sub> - 3)<sup>2</sup> + (<span>y</span><sub>1</sub> + 2)<sup>2</sup> = 5<span>r</span><sup>2</sup></p><p><strong>Step 2:</strong> Length of tangent from P to circle <span>(x</span> - 3)<sup>2</sup> + (<span>y</span> + 2)<sup>2</sup> = <span>r</span><sup>2</sup> is:</p><p>\[\sqrt{5r^2 - r^2} = \sqrt{4r^2} = 2r\]</p><p><strong>Step 3:</strong> From the tangent length equation: \[\sqrt{16} = 2r \Rightarrow r = 8\]</p><p><strong>Step 4:</strong> Area between two circles = \[\pi \cdot 5r^2 - \pi r^2 = 4\pi r^2 = 4\pi \times 64 = 256\pi\]</p><p>∴ <span>k</span> = 256</p>
Correct Answer: 256