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: Note the given coordinates.
Center $O(-5, 3)$, $P(3, 1)$, $Q(0, 6)$. PQ is given as tangent to the circle at Q. We need to show $PQ \perp OQ$.
Step 2: Calculate PQ using the Distance Formula.
\[ PQ^2 = (0 - 3)^2 + (6 - 1)^2 = (-3)^2 + (5)^2 = 9 + 25 = 34 \]
Step 3: Calculate OQ using the Distance Formula.
\[ OQ^2 = (0 - (-5))^2 + (6 - 3)^2 = (5)^2 + (3)^2 = 25 + 9 = 34 \]
Step 4: Calculate OP using the Distance Formula.
\[ OP^2 = (3 - (-5))^2 + (1 - 3)^2 = (8)^2 + (-2)^2 = 64 + 4 = 68 \]
Step 5: Apply the Converse of Pythagoras theorem.
Check if $OQ^2 + PQ^2 = OP^2$: $34 + 34 = 68$ = $OP^2$. This confirms $\angle OQP = 90^\circ$, i.e., $PQ \perp OQ$.
Step 6: Conclusion.
Since $OQ^2 + PQ^2 = OP^2$, by the converse of the Pythagorean theorem, $PQ \perp OQ$. Hence proved.
\[ \boxed{PQ \perp OQ \text{ (proved)}} \]