Let the lines $y+2x=\sqrt{11}+7\sqrt{7}$ and $2y+x=2\sqrt{11}+6\sqrt{7}$ be normal to a circle $C:(x-h)^{2}+(y-k)^{2}=r^{2}$. If the line $\sqrt{11}y-3x=\frac{5\sqrt{77}}{3}+11$ is tangent to the circle C, then the value of $(5h-8k)^{2}+5r^{2}$ is equal to ____.

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Circle

Let the lines

y + 2 x = 11 + 77

and

2 y + x = 211 + 67

be normal to a circle

C : (x - h)^{2} + (y - k)^{2} = r^{2}

. If the line

11 y - 3 x = \frac{5 77}{3} + 11

is tangent to the circle C, then the value of

(5 h - 8 k)^{2} + 5 r^{2}

is equal to ____.

Answer: 816

The center of the circle is at the intersection of the two normals:

y + 2 x = 11 + 77

2 y + x = 211 + 67

Solving: multiply the first by 2:

2 y + 4 x = 211 + 147

Subtract the second:

3 x = 87

, so

x = \frac{8 7}{3}

Substituting back:

y = 11 + \frac{5 7}{3}

So the center is

(h, k) = (\frac{8 7}{3}, 11 + \frac{5 7}{3})

.

The radius is the perpendicular distance from the center to the tangent

11 y - 3 x = \frac{5 77}{3} + 11

r = \frac{11 ( 11 + \frac{5 7}{3} ) - 3 ( \frac{8 7}{3} ) - \frac{5 77}{3} - 11}{11 + 9} = 4 \frac{7}{5}

Therefore:

5 h - 8 k = 5 (\frac{8 7}{3}) - 8 (11 + \frac{5 7}{3}) = 0 - 811

(5 h - 8 k)^{2} + 5 r^{2} = 64 (11) + 5 (16 \cdot \frac{7}{5}) = 704 + 112 = 816

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Integer answer