Question

In the given figure, PQ is a tangent to a circle with centre \(O(-5, 3)\). If coordinates of P and Q are \((3, 1)\) and \((0, 6)\) respectively, then using distance formula, show that \(PQ \perp OQ\).

Hint:

Using \(d^2\) directly instead of \(d\) is much faster and cleaner because you avoid writing square roots and carrying them throughout the calculation.

Answer 1

Correct Answer: 4

Step 1: Understanding the Question:
We are given a circle with center \(O(-5, 3)\) and a tangent line \(PQ\) where \(P(3, 1)\) and \(Q(0, 6)\) are points.
We need to prove that \(PQ\) is perpendicular to \(OQ\) using the distance formula.

Step 2: Key Formula or Approach:
1. Distance Formula:
\[ d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 \]
2. Converse of Pythagoras' Theorem:
If in a triangle \(\triangle OQP\), the sum of the squares of two sides is equal to the square of the longest side:
\[ OP^2 = OQ^2 + PQ^2 \]
then the triangle is right-angled at \(Q\), which implies \(PQ \perp OQ\).

Step 3: Detailed Explanation:

• Let us calculate the squared distance of \(OQ^2\) using points \(O(-5, 3)\) and \(Q(0, 6)\):
\[ OQ^2 = (0 - (-5))^2 + (6 - 3)^2 \] \[ OQ^2 = (5)^2 + (3)^2 = 25 + 9 = 34 \]

• Let us calculate the squared distance of \(PQ^2\) using points \(P(3, 1)\) and \(Q(0, 6)\):
\[ PQ^2 = (0 - 3)^2 + (6 - 1)^2 \] \[ PQ^2 = (-3)^2 + (5)^2 = 9 + 25 = 34 \]

• Let us calculate the squared distance of \(OP^2\) using points \(O(-5, 3)\) and \(P(3, 1)\):
\[ OP^2 = (3 - (-5))^2 + (1 - 3)^2 \] \[ OP^2 = (8)^2 + (-2)^2 = 64 + 4 = 68 \]

• Now, check the relation between the squared lengths:
- Sum of the squares of the two shorter sides:
\[ OQ^2 + PQ^2 = 34 + 34 = 68 \] - This is exactly equal to the square of the longest side:
\[ OP^2 = 68 \] - Since \(OP^2 = OQ^2 + PQ^2\), by the converse of Pythagoras' theorem, \(\triangle OQP\) is right-angled at vertex \(Q\).
- Therefore, the segment \(PQ\) is perpendicular to \(OQ\).


Step 4: Final Answer:
Hence, \(PQ \perp OQ\) is proved using the distance formula.