C₁ is the circle with equation
$$(x - 7)^{2} + (y + 5)^{2} = 100.$$
(a)
(i) State the centre and radius of C₁ - Scottish Highers Maths - Question 14 - 2022
Question 14
C₁ is the circle with equation
$$(x - 7)^{2} + (y + 5)^{2} = 100.$$
(a)
(i) State the centre and radius of C₁.
(ii) Hence, or otherwise, show that the point P... show full transcript
Worked Solution & Example Answer:C₁ is the circle with equation
$$(x - 7)^{2} + (y + 5)^{2} = 100.$$
(a)
(i) State the centre and radius of C₁ - Scottish Highers Maths - Question 14 - 2022
Step 1
State the centre and radius of C₁.
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Answer
The equation of the circle C₁ is given as
(x−7)2+(y+5)2=100.
From this equation, we can determine that the center of C₁ is at the point (7, -5) and the radius is given by the square root of 100.
Thus, the radius is ( r = 10 ).
Step 2
Hence, or otherwise, show that the point P(−2, 7) lies outside C₁.
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Answer
To verify whether the point P(−2, 7) lies inside, on, or outside the circle C₁, we first calculate the distance ( d ) from point P to the center of C₁ (7, -5). The distance is calculated using the distance formula: d=(x2−x1)2+(y2−y1)2
Substituting the coordinates gives: d=((−2)−7)2+(7+5)2=(−9)2+(12)2=81+144=225=15.
Since the radius of C₁ is 10, and ( d = 15 > 10 ), this shows that the point P(−2, 7) lies outside circle C₁.
Step 3
Determine the value(s) of r for which circles C₁ and C₂ have exactly one point of intersection.
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Answer
For circles C₁ and C₂ to intersect at exactly one point, the distance between their centers must equal the sum or the difference of their radii.
We already calculated the distance from C₁'s center to point P: ( d = 15 ).
Let the radius of C₂ be ( r ). The conditions for exactly one intersection point are:
( d = r + 10 )
( d = |r - 10| )
From the first condition:
\Rightarrow r = 5. $$
From the second condition:
$$ 15 = r - 10
\Rightarrow r = 25. $$
Thus, the values of \( r \) for which circles C₁ and C₂ intersect at exactly one point are: \( r = 5 \) and \( r = 25. \)
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