Exercise 10.1 answers: Conic Sections

Open NCERT p. 181Matches NCERT’s answer

1. A circle's equation says every point (x, y) on it is the fixed distance r from the centre.
2. Here centre is (0, 2) and radius is 2, so (x − 0)² + (y − 2)² = 2².
3. Multiply out: x² + y² − 4y + 4 = 4.
4. Simplify: x² + y² − 4y = 0.

Answerx² + y² − 4y = 0

Watch this explained “Put it on axes”, 1:42 into Fixed distance from a fixed point, turned into an equation · हिंदी में देखें

Open NCERT p. 181Matches NCERT’s answer

1. Use (x − h)² + (y − k)² = r² with centre (−2, 3) and radius 4.
2. Since h is −2, the bracket x − h becomes x + 2, so (x + 2)² + (y − 3)² = 16.
3. Multiply out: x² + 4x + 4 + y² − 6y + 9 = 16.
4. Simplify: x² + y² + 4x − 6y − 3 = 0.

Answerx² + y² + 4x − 6y − 3 = 0

Watch this explained “The sign that catches everyone”, 3:35 into Fixed distance from a fixed point, turned into an equation · हिंदी में देखें

Open NCERT p. 181Matches NCERT’s answer

1. Use (x − h)² + (y − k)² = r² with centre (1/2, 1/4) and radius 1/12.
2. This gives (x − 1/2)² + (y − 1/4)² = (1/12)² = 1/144.
3. Multiply out: x² − x + 1/4 + y² − y/2 + 1/16 = 1/144.
4. Collect the numbers over 144: 1/4 + 1/16 − 1/144 = 36/144 + 9/144 − 1/144 = 44/144 = 11/36.
5. So x² + y² − x − y/2 + 11/36 = 0. Multiplying every term by 36 gives the same circle without fractions: 36x² + 36y² − 36x − 18y + 11 = 0.

Answerx² + y² − x − y/2 + 11/36 = 0, that is, 36x² + 36y² − 36x − 18y + 11 = 0

Watch this explained “Put it on axes”, 1:42 into Fixed distance from a fixed point, turned into an equation · हिंदी में देखें

Open NCERT p. 181Matches NCERT’s answer

1. Use (x − h)² + (y − k)² = r² with centre (1, 1) and radius √2.
2. Write (x − 1)² + (y − 1)² = (√2)² = 2.
3. Multiply out: x² − 2x + 1 + y² − 2y + 1 = 2.
4. Simplify: x² + y² − 2x − 2y = 0.

Answerx² + y² − 2x − 2y = 0

Watch this explained “Put it on axes”, 1:42 into Fixed distance from a fixed point, turned into an equation · हिंदी में देखें

Open NCERT p. 181Matches NCERT’s answer

1. Use (x − h)² + (y − k)² = r² with centre (−a, −b) and radius √(a² − b²).
2. Since h = −a and k = −b, the brackets become x + a and y + b: (x + a)² + (y + b)² = a² − b².
3. Multiply out: x² + 2ax + a² + y² + 2by + b² = a² − b².
4. The a² on both sides cancels, leaving x² + y² + 2ax + 2by + 2b² = 0.

Answerx² + y² + 2ax + 2by + 2b² = 0

Watch this explained “The sign that catches everyone”, 3:35 into Fixed distance from a fixed point, turned into an equation · हिंदी में देखें

Open NCERT p. 181Matches NCERT’s answer

1. Compare (x + 5)² + (y − 3)² = 36 with the template (x − h)² + (y − k)² = r².
2. Writing x + 5 as x − (−5) shows h = −5.
3. The y-bracket already matches the template, so k = 3.
4. r² = 36, so r = 6.

AnswerCentre (−5, 3), radius 6.

Watch this explained “The sign that catches everyone”, 3:35 into Fixed distance from a fixed point, turned into an equation · हिंदी में देखें

Open NCERT p. 181Matches NCERT’s answer

1. Group the x-terms and y-terms: (x² − 4x) + (y² − 8y) = 45.
2. Complete the square on each: (x² − 4x + 4) + (y² − 8y + 16) = 45 + 4 + 16.
3. Write as squares: (x − 2)² + (y − 4)² = 65.
4. So the centre is (2, 4) and the radius is √65.

AnswerCentre (2, 4), radius √65.

Watch this explained “Reading it backwards”, 5:52 into Fixed distance from a fixed point, turned into an equation · हिंदी में देखें

Open NCERT p. 181Matches NCERT’s answer

1. Group the x-terms and y-terms: (x² − 8x) + (y² + 10y) = 12.
2. Complete the square on each: (x² − 8x + 16) + (y² + 10y + 25) = 12 + 16 + 25.
3. Write as squares: (x − 4)² + (y + 5)² = 53.
4. So the centre is (4, −5) and the radius is √53.

AnswerCentre (4, −5), radius √53.

Watch this explained “Reading it backwards”, 5:52 into Fixed distance from a fixed point, turned into an equation · हिंदी में देखें

Open NCERT p. 181Matches NCERT’s answer

1. Divide the whole equation by 2 so both squared terms have coefficient 1: x² + y² − x/2 = 0.
2. Complete the square on x: (x² − x/2 + 1/16) + y² = 1/16.
3. Write as squares: (x − 1/4)² + y² = 1/16.
4. So the centre is (1/4, 0) and the radius is 1/4.

AnswerCentre (1/4, 0), radius 1/4.

Watch this explained “Where the completion refuses”, 7:50 into Fixed distance from a fixed point, turned into an equation · हिंदी में देखें

Open NCERT p. 181Matches NCERT’s answer

1. Let the centre be (h, k). Both (4, 1) and (6, 5) are on the circle, so they are the same distance from the centre: (h − 4)² + (k − 1)² = (h − 6)² + (k − 5)².
2. Multiply out; h² and k² cancel: −8h + 16 − 2k + 1 = −12h + 36 − 10k + 25. This simplifies to 4h + 8k = 44, that is, h + 2k = 11.
3. The centre is on the line 4x + y = 16, so 4h + k = 16.
4. From h + 2k = 11, h = 11 − 2k. Then 4(11 − 2k) + k = 16, so 44 − 7k = 16, giving k = 4 and h = 3.
5. r² = (4 − 3)² + (1 − 4)² = 1 + 9 = 10.
6. The circle is (x − 3)² + (y − 4)² = 10, that is, x² − 6x + 9 + y² − 8y + 16 = 10, so x² + y² − 6x − 8y + 15 = 0.

Answerx² + y² − 6x − 8y + 15 = 0

Watch this explained “Three unknowns, three conditions”, 9:58 into Fixed distance from a fixed point, turned into an equation · हिंदी में देखें

Open NCERT p. 181Matches NCERT’s answer

1. Let the centre be (h, k). Both (2, 3) and (−1, 1) are on the circle, so (h − 2)² + (k − 3)² = (h + 1)² + (k − 1)².
2. Multiply out; h² and k² cancel: −4h + 4 − 6k + 9 = 2h + 1 − 2k + 1. This simplifies to 6h + 4k = 11.
3. The centre is on x − 3y − 11 = 0, so h − 3k = 11, that is, h = 3k + 11.
4. Substitute: 6(3k + 11) + 4k = 11, so 22k + 66 = 11, giving k = −55/22 = −5/2, and h = 3(−5/2) + 11 = 7/2.
5. r² = (7/2 − 2)² + (−5/2 − 3)² = (3/2)² + (−11/2)² = 9/4 + 121/4 = 130/4 = 65/2.
6. The circle is (x − 7/2)² + (y + 5/2)² = 65/2. Multiply out: x² − 7x + 49/4 + y² + 5y + 25/4 = 65/2, and 49/4 + 25/4 − 130/4 = −56/4 = −14, so x² + y² − 7x + 5y − 14 = 0.

Answerx² + y² − 7x + 5y − 14 = 0

Watch this explained “Three unknowns, three conditions”, 9:58 into Fixed distance from a fixed point, turned into an equation · हिंदी में देखें

Open NCERT p. 181Matches NCERT’s answer

1. The centre lies on the x-axis, so call it (h, 0).
2. The circle passes through (2, 3), and its radius is 5, so (h − 2)² + (0 − 3)² = 5².
3. Solve: (h − 2)² = 16, so h = 6 or h = −2 — two centres fit the conditions.
4. The two circles are (x − 6)² + y² = 25 and (x + 2)² + y² = 25, i.e. x² + y² − 12x + 11 = 0 and x² + y² + 4x − 21 = 0.

Answerx² + y² − 12x + 11 = 0 or x² + y² + 4x − 21 = 0

Watch this explained “When three conditions leave two”, 10:59 into Fixed distance from a fixed point, turned into an equation · हिंदी में देखें

Open NCERT p. 181Matches NCERT’s answer

1. Making intercepts a and b means the circle also passes through (a, 0) and (0, b), besides (0, 0).
2. Take the general circle x² + y² + 2gx + 2fy + c = 0. Through (0, 0), c = 0.
3. Through (a, 0): a² + 2ga = 0, so g = −a/2. Through (0, b): b² + 2fb = 0, so f = −b/2.
4. So the equation is x² + y² − ax − by = 0.

Answerx² + y² − ax − by = 0

Watch this explained “Three unknowns, three conditions”, 9:58 into Fixed distance from a fixed point, turned into an equation · हिंदी में देखें

Open NCERT p. 181Matches NCERT’s answer

1. The radius is the distance from the centre (2, 2) to the point (4, 5) on the circle.
2. r² = (4 − 2)² + (5 − 2)² = 4 + 9 = 13.
3. Use (x − 2)² + (y − 2)² = 13.
4. Expand and simplify: x² + y² − 4x − 4y − 5 = 0.

Answerx² + y² − 4x − 4y − 5 = 0

Watch this explained “Put it on axes”, 1:42 into Fixed distance from a fixed point, turned into an equation · हिंदी में देखें

Open NCERT p. 181Matches NCERT’s answer

1. The circle x² + y² = 25 has centre (0, 0) and r² = 25.
2. Find the squared distance of the point from the centre: (−2.5)² + (3.5)² = 6.25 + 12.25 = 18.5.
3. Compare 18.5 with 25: it is smaller.
4. So the point lies inside the circle.

Answerinside

Watch this explained “Inside, outside, or on”, 12:04 into Fixed distance from a fixed point, turned into an equation · हिंदी में देखें