Math Linear Algebra help

Question

Find equations for all the planes that intersect the y-axis at y = 1 and the z-axis at z = 2, and are tangent to the sphere (𝑥 − 2)^2 + 𝑦^2 + 𝑧^2 = 4. You cannot use calculus

Please help and if you can show steps that would be greatly appreciated, thank you!

Answer 1

Points (0,1,0) and (0, 0, 2) belong to tangent plane ax + by + cz + d = 0.

We have 2 equations: b + d = 0 and 2c + d = 0. So, b = - d and c = - d/2.

Equation of tangent plane now ax - dy - d/2z + d = 0. Distanse from center (2, 0, 0) to tangent plane equel radius r = 2. Normal equation of plane now: (ax - dy -d/2z +d)/√(a² + d² + d²/4).

Then distance from center (2, 0, 0) to the tangent plane is I2a + dI/√(a² + 5/4d²) = 2;

Raising to the square gives: 4a² + 4ad + d² = 4a² + 5d²; 4ad = 4d²; a = d.

So, equation of tangent plane: dx - dy- d/2z + d = 0; or redused by d/2 we get: 2x - 2y - z + 2 = 0