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Question: Find the equation of the largest circle passing through the point (1, 1) and (2, 2) and which does n...

Find the equation of the largest circle passing through the point (1, 1) and (2, 2) and which does not cross the boundaries of the first quadrant.

A

(x-1)^2 + (y-2)^2 = 1

B

(x-2)^2 + (y-1)^2 = 1

C

x^2 + y^2 - 2x - 4y + 4 = 0

D

x^2 + y^2 - 4x - 2y + 4 = 0

Answer

The largest circles are (x−1)2+(y−2)2=1(x-1)^2 + (y-2)^2 = 1 and (x−2)2+(y−1)2=1(x-2)^2 + (y-1)^2 = 1.

Explanation

Solution

Let the equation of the circle be (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2. The condition that the circle does not cross the boundaries of the first quadrant implies h−r≥0h-r \ge 0 and k−r≥0k-r \ge 0, so h≥rh \ge r and k≥rk \ge r.

Since the circle passes through (1, 1) and (2, 2): (1−h)2+(1−k)2=r2(1-h)^2 + (1-k)^2 = r^2 (1) (2−h)2+(2−k)2=r2(2-h)^2 + (2-k)^2 = r^2 (2)

Equating (1) and (2): (1−h)2+(1−k)2=(2−h)2+(2−k)2(1-h)^2 + (1-k)^2 = (2-h)^2 + (2-k)^2 1−2h+h2+1−2k+k2=4−4h+h2+4−4k+k21 - 2h + h^2 + 1 - 2k + k^2 = 4 - 4h + h^2 + 4 - 4k + k^2 2−2h−2k=8−4h−4k2 - 2h - 2k = 8 - 4h - 4k 2h+2k=62h + 2k = 6 h+k=3  ⟹  k=3−hh + k = 3 \implies k = 3-h

Substitute k=3−hk=3-h into (1): (1−h)2+(1−(3−h))2=r2(1-h)^2 + (1-(3-h))^2 = r^2 (1−h)2+(h−2)2=r2(1-h)^2 + (h-2)^2 = r^2 1−2h+h2+h2−4h+4=r21 - 2h + h^2 + h^2 - 4h + 4 = r^2 r2=2h2−6h+5r^2 = 2h^2 - 6h + 5

The constraints h≥rh \ge r and k≥rk \ge r become h≥rh \ge r and 3−h≥r3-h \ge r. Squaring these (since r>0r > 0): r2≤h2r^2 \le h^2 and r2≤(3−h)2r^2 \le (3-h)^2.

Substituting r2=2h2−6h+5r^2 = 2h^2 - 6h + 5: 2h2−6h+5≤h2  ⟹  h2−6h+5≤0  ⟹  (h−1)(h−5)≤0  ⟹  1≤h≤52h^2 - 6h + 5 \le h^2 \implies h^2 - 6h + 5 \le 0 \implies (h-1)(h-5) \le 0 \implies 1 \le h \le 5. 2h2−6h+5≤(3−h)2  ⟹  2h2−6h+5≤9−6h+h2  ⟹  h2−4≤0  ⟹  (h−2)(h+2)≤0  ⟹  −2≤h≤22h^2 - 6h + 5 \le (3-h)^2 \implies 2h^2 - 6h + 5 \le 9 - 6h + h^2 \implies h^2 - 4 \le 0 \implies (h-2)(h+2) \le 0 \implies -2 \le h \le 2.

Combining these, we get 1≤h≤21 \le h \le 2.

We want to maximize r2=2h2−6h+5r^2 = 2h^2 - 6h + 5 for h∈[1,2]h \in [1, 2]. This is a parabola opening upwards. The maximum on a closed interval occurs at the endpoints. At h=1h=1: r2=2(1)2−6(1)+5=2−6+5=1r^2 = 2(1)^2 - 6(1) + 5 = 2 - 6 + 5 = 1. At h=2h=2: r2=2(2)2−6(2)+5=8−12+5=1r^2 = 2(2)^2 - 6(2) + 5 = 8 - 12 + 5 = 1.

The maximum r2r^2 is 1, so the maximum radius r=1r=1.

Case 1: h=1h=1 k=3−h=3−1=2k = 3-h = 3-1 = 2. Center is (1, 2), radius is 1. Equation: (x−1)2+(y−2)2=1(x-1)^2 + (y-2)^2 = 1. Check constraints: h=1≥r=1h=1 \ge r=1 (True), k=2≥r=1k=2 \ge r=1 (True).

Case 2: h=2h=2 k=3−h=3−2=1k = 3-h = 3-2 = 1. Center is (2, 1), radius is 1. Equation: (x−2)2+(y−1)2=1(x-2)^2 + (y-1)^2 = 1. Check constraints: h=2≥r=1h=2 \ge r=1 (True), k=1≥r=1k=1 \ge r=1 (True).

Both equations represent the largest circles satisfying the given conditions.