{"id":35508,"date":"2022-09-19T17:09:55","date_gmt":"2022-09-19T11:39:55","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35508"},"modified":"2022-09-19T17:09:55","modified_gmt":"2022-09-19T11:39:55","slug":"inter-2nd-year-maths-2b-circle-solutions-ex-1d","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/inter-2nd-year-maths-2b-circle-solutions-ex-1d\/","title":{"rendered":"Inter 2nd Year Maths 2B Circle Solutions Ex 1(d)"},"content":{"rendered":"

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions<\/a> Inter 2nd Year Maths 2B Circle Solutions Exercise 1(d) will help students to clear their doubts quickly.<\/p>\n

Intermediate 2nd Year Maths 2B Circle Solutions Exercise 1(d)<\/h2>\n

I.<\/span><\/p>\n

Question 1.
\nFind the condition that the tangents drawn from (0, 0) to S \u2261 x\u00b2 + y\u00b2 + 2gx + 2fy + c = 0 be perpendicular to each other.
\nSolution:
\nIf \u03b8 is the angle between the pair of tangents from P1<\/sub> then tan \\(\\frac{\\theta}{2}=\\frac{r}{\\sqrt{s_{11}}}\\)
\n\"Inter
\ng\u00b2 + f\u00b2 – c = c
\ng\u00b2 + f\u00b2 = 2c
\nThis is the required condition.<\/p>\n

Question 2.
\nFind the chord of contact of (0, 5) with respect to the circle x\u00b2 + y\u00b2 – 5x + 4y – 2 = 0
\nSolution:
\nEquation of the chord of contact is S1<\/sub> = 0
\ni.e., xx1<\/sub> + yy1<\/sub> + g(x + x1<\/sub>) + f(y + y1<\/sub>) + c = 0
\nEquation of the circle is
\nS = x\u00b2 + y\u00b2 – 5x + 4y – 2 = 0
\nEquation of the chord of contact is
\nx. 0 + y. 5 – \\(\\frac{5}{2}\\) (x + 0) + 2(y + 5) – 2 = 0
\nMultiplying with 2
\n10y – 5x + 4y + 20 – 4 = 0 – 5x + 14y +16 = 0
\nor 5x – 14y – 16 = 0<\/p>\n

Question 3.
\nFind the chord of contact of (1, 1) to the circle x\u00b2 + y\u00b2 = 9.
\nSolution:
\nEquation of the circle is x\u00b2 + y\u00b2 = 9
\nEquation of the chord of contact is
\nx.1 + y. 1 =9
\ni.e. x + y = 9<\/p>\n

\"Inter<\/p>\n

Question 4.
\nFind the polar of (1, 2) with respect to x\u00b2 + y\u00b2 = 7.
\nSolution:
\nPolar of P(x1<\/sub>, y1<\/sub>) w.r. to S = 0
\nxx1<\/sub> + yy1<\/sub> = a\u00b2
\nx + 2y – 7 = 0 is the polar equation.<\/p>\n

Question 5.
\nFind the polar of (3, -1) with respect to 2x\u00b2 + 2y\u00b2 = 11.
\nSolution:
\nEquation of circle is 2x\u00b2 + 2y\u00b2 = 11
\nx\u00b2 + y\u00b2 = \\(\\frac{11}{2}\\)
\nEquation of polar is xx1<\/sub> + yy1<\/sub> = a\u00b2
\nx(3) + y(-1) = \\(\\frac{11}{2}\\)
\nEquation-of polar is 6x – 2y -11 = 0<\/p>\n

Question 6.
\nFind the polar of (1, -2) with respect to x\u00b2 + y\u00b2 – 10x – 10y + 25 = 0
\nSolution:
\nEquation of the circle is
\nx\u00b2 + y\u00b2 – 10x – 10y + 25 = 0
\nEquation of the polar is S1<\/sub> = 0
\nPolar of P(1, -2) is
\nx .1 + y(-2) – 5(x + 1) – 5(y – 2) + 25 = 0
\n\u21d2 x – 2y – 5x – 5 – 5y + 10 + 25 = 0
\n\u21d2 -4x – 7y + 30 = 0
\n\u2234 4x + 7y – 30 = 0<\/p>\n

Question 7.
\nFind the pole of ax + by + c = 0 (c + 0) with respect to x\u00b2 + y\u00b2 = r\u00b2.
\nSolution:
\nLet (x1<\/sub>, y1<\/sub>) be pole, the polar equation be
\nxx1<\/sub> + yy1<\/sub> – r\u00b2 = 0 ………… (i)
\nax + by + c = 0 ………….. (ii)
\n(i) and (ii) both are same (polar) equations.
\n\"Inter<\/p>\n

Question 8.
\nFind the pole of 3x + 4y – 45 = 0 with respect to x\u00b2 + y\u00b2 – 6x – 8y + 5 = 0.
\nSolution:
\nEquation of polar is
\nxx1<\/sub> + yy1<\/sub> – 3(x + x1<\/sub>) – 4(y + y1<\/sub>) + 5 = 0
\nx(x1<\/sub> – 3) + y(y1<\/sub> – 4) – 3x1<\/sub> – 4y1<\/sub> + 5 = 0 ……….. (i)
\nPolar equation is same 3x + 4y – 45 = 0 ……….. (ii)
\nComparing (i) and (ii) we get
\n\"Inter
\nPole is (6, 8).<\/p>\n

Question 9.
\nFind the pole of x – 2y + 22 = 0 with respect to x\u00b2 + y\u00b2 – 5x + 8y + 6 = 0.
\nSolution:
\nPolar equation is
\nxx1<\/sub> + yy1<\/sub> – \\(\\frac{5}{2}\\)(x + x1<\/sub>) + 4(y + y1<\/sub>) + 6 = 0
\n(or) x(x1<\/sub> – \\(\\frac{5}{2}\\)) + y(y1<\/sub> + 4) – \\(\\frac{5}{2}\\)
\nx1<\/sub> + 4y1<\/sub> + 6 = 0 …………. (i)
\nx – 2y + 22 = 0 ………….. (ii)
\n(i) and (ii) both are equations of the same line comparing it we get
\n\"Inter
\n\"Inter
\ny1<\/sub> = -3
\n\u2234 Pole is (2, -3)<\/p>\n

Question 10.
\nShow that the points (-6, 1) and (2, 3) are conjugate points with respect to the circle x\u00b2 + y\u00b2 – 2x + 2y + 1 =0.
\nSolution:
\nPolar equation w.r.t. (2, 3) is
\n2x + 3y – 1(x + 2) + 1(y + 3) + 1 = 0
\nx + 4y + 2 = 0 ………. (i)
\nNow (-6, 1) should satisfy equation (i) if it is conjugate point
\n(-6) + 4(1) + 2 = 0
\n\u2234 (-6, 1) and (2, 3) are conjugate w.r.t. circle.<\/p>\n

\"Inter<\/p>\n

Question 11.
\nShow that the points (4, 2) (3, -5) are conjugate points with respect to the circle x\u00b2 + y\u00b2 – 3x – 5y + 1 = 0.
\nSolution:
\nPolar equation w.r.t. (4, 2) is
\n4x + 2y – \\(\\frac{3}{2}\\)(x + 4) – \\(\\frac{5}{2}\\)(y + 2) + 1 = 0 ………. (i)
\nNow (3, -5) should satisfy equation (i) if it is conjugate points.
\n4(3) – 5 \u00d7 2 – \\(\\frac{3}{2}\\)(3 + 4) – \\(\\frac{5}{2}\\)(-5 + 2) + 1
\n12 – 10 – \\(\\frac{21}{2}\\) + \\(\\frac{15}{2}\\) + 1 = 0
\nHence given points are conjugate.<\/p>\n

Question 12.
\nFind the value of k if kx + 3y – 1 = 0, 2x + y + 5 = 0 are conjugate lines with respect to the circle x\u00b2 + y\u00b2 – 2x – 4y – 4 = 0.
\nSolution:
\nLet pole be (x1<\/sub>, y1<\/sub>) polar equation w.r.t.
\nx\u00b2 + y\u00b2 – 2x – 4y – 4 = 0 be
\nxx1<\/sub> + yy1<\/sub> -1 (x + x1<\/sub>) – 2(y + y1<\/sub>) – 4 = 0
\nx(x1<\/sub> – 1) + y(y1<\/sub> – 2) – x1<\/sub> – 2y1<\/sub> – 4 = 0 …….. (i)
\n2x + y + 5 = 0 is polar so comparing with equation (i) we get
\n\"Inter
\nx1<\/sub> = -1, y1<\/sub> = 1 Pole (-1, 1)
\nkx + 3y – 1 = 0 is polar so it should satisfy
\n(1, 1)
\nk(-1) + 3(1) – 1 = 0
\n-k + 2 = 0
\nk = 2<\/p>\n

Question 13.
\nFind the value of k if x + y – 5 = 0, 2x + ky – 8 = 0 are conjugate with respect to the circle x\u00b2 + y\u00b2 – 2x – 2y – 1 = 0.
\nSolution:
\nIf l1<\/sub>x + m1<\/sub>y + n1<\/sub> = 0; l2<\/sub>x + m2<\/sub>y + n2<\/sub> = 0 are conjugate, S = 0 then,
\nr\u00b2(l1<\/sub>l2<\/sub> + m1<\/sub>m2<\/sub>)
\n= (l1<\/sub>g + m1<\/sub>f – n1<\/sub>) (l2<\/sub>g + m2<\/sub>f – n2<\/sub>)
\nl1<\/sub> = 1, m1<\/sub> = 1, n1<\/sub> = -5
\nl2<\/sub> = 2, m2<\/sub> = k, n2<\/sub> = – 8
\ng = -1, f = -1, r\u00b2 = 3
\n\u2234 3(1.2 + k) = (-1 -1 + 5) (-2 – k + 8)
\n6k = 18 – 6 = 12 \u21d2 k = 2<\/p>\n

Question 14.
\nFind the value of k if the points (1, 3) and (2, k) are conjugate with respect to the circle x\u00b2 + y\u00b2 = 35.
\nSolution:
\nEquation of the circle is x\u00b2 + y\u00b2 = 35
\nPolar of P(1, 3) is x. 1 + y. 3 = 35
\nx + 3y = 35
\nP(1, 3) and Q(2, k) are conjugate points The polar of P passes through Q
\n2 + 3k = 35
\n3k = 33
\nk = 11<\/p>\n

Question 15.
\nFind the value of k if the points (4, 2) and (k, -3) are conjugate points with respect to the circle x\u00b2 + y\u00b2 – 5x + 8y + 6 = 0.
\nSolution:
\nEquation of the circle is x\u00b2 + y\u00b2 – 5x + 8y + 6 = 0
\nPolar of P(4, 2) is
\nx.4 + y.2 – \\(\\frac{5}{2}\\)(x + 4) + 4(y + 2) + 6 = 0
\n8x + 4y – 5x – 20 + 8y + 16 + 12 = 0
\n3x + 12y + 8 = 0
\nP(4, 2), Q(k, -3) are conjugate points
\nPolar of P passes through Q
\n\u2234 3k – 36 + 8 = 0
\n3k = 28 \u21d2 k = \\(\\frac{28}{2}\\)<\/p>\n

II.<\/span><\/p>\n

Question 1.
\nFind the angle between the tangents drawn from (3, 2) to the circle x\u00b2 + y\u00b2 – 6x + 4y – 2 = 0.
\nSolution:
\nEquation of the circle is
\nS \u2261 x\u00b2 + y\u00b2 – 6x + 4y – 2 = 0
\n\"Inter
\nAngle between the tangent at P = cos-1<\/sup> (\\(\\frac{7}{8}\\))<\/p>\n

\"Inter<\/p>\n

Question 2.
\nFind the angle between the pair of tangents dravyn from (1, 3) to the circle x\u00b2 + y\u00b2 – 2x + 4y – 11 = 0.
\nSolution:
\nEquation of the circle is
\nS1<\/sub> \u2261 x\u00b2 + y\u00b2 – 2x + 4y – 11 = 0
\nr = \\(\\sqrt{1+4+11}\\) = 4
\nS11<\/sub> =1 +9-2 + 12-11 = 9
\n\"Inter
\nAngle between the tangents = cos-1<\/sup> (\\(\\frac{7}{25}\\))<\/p>\n

Question 3.
\nFind the angle between the pair of tangents drawn from (0,0) to the circle x\u00b2 + y\u00b2 – 14x + 2y + 25 = 0.
\nSolution:
\nEquation of the circle is
\nx\u00b2 + y\u00b2 – 14x + 2y + 25 = 0
\nr = \\(\\sqrt{49+1-25}\\) = 5
\nS11<\/sub> = 0 + 0 – 0 + 0 + 25 = 25 \u21d2 \u221aS11<\/sub> = 5
\ntan \u03b8 = \\(\\frac{r}{\\sqrt{s_{11}}}\\) = \\(\\frac{5}{5}\\) = 1 \u21d2 \u03b8 = \\(\\frac{\\pi}{4}\\)
\nAngle between the tangents = 2\u03b8 = \\(\\frac{\\pi}{2}\\)<\/p>\n

Question 4.
\nFind the locus of P if the tangents drawn from P to x\u00b2 + y\u00b2= a\u00b2 include an angle \u03b1.
\nSolution:
\nEquation of the circle is x\u00b2 + y\u00b2 = a\u00b2
\nr = radius = a
\nSuppose \u03b8 (x1<\/sub>, y1<\/sub>) is any point as the locus
\n\"Inter
\n\"Inter<\/p>\n

Question 5.
\nFind the locus of P if the tangents drawn from P to x2 + y2= a2 are perpendicular to each other.
\nSolution:
\nEquation of the circle is x\u00b2 + y\u00b2 = a\u00b2
\nr = a
\nLet P(x1<\/sub>, y1<\/sub>) be any point on the locus
\nS11<\/sub> = x\u00b21<\/sub> + y\u00b21<\/sub> – a\u00b2
\n\"Inter
\nSquaring and cross-multiplying
\na\u00b2 = x\u00b2 + y\u00b2 – a\u00b2
\nx\u00b2 + y\u00b2 = 2a\u00b2
\nLocus of P (x1<\/sub>, y1<\/sub>) is x\u00b2 + y\u00b2= 2a\u00b2<\/p>\n

Question 6.
\nFind the slope of the polar of (1, 3) with respect to the circle x\u00b2 + y\u00b2 – 4x – 4y – 4= 0. Also find the distance from the centre to it.
\nSolution:
\nEquation of the circle is x\u00b2 + y\u00b2 – 4x – 4y – 4 =0
\nPolar of P (1, 3) is
\nx.1 + y.3 – 2(x + 1) – 2(y + 3) – 4 = 0
\nx + 3y – 2x – 2 – 2y – 6 – 4 = 0
\n-x + y – 12 = 0
\nSlope of the polar is – \\(\\frac{a}{b}=\\frac{1}{1}\\) = 1
\nDistance from the centre
\n\"Inter<\/p>\n

Question 7.
\nIf ax + by + c = 0 is the polar of (1, 1) with respect to the circle x\u00b2 + y\u00b2 – 2x + 2y +1=0 and H.C.F. of a, b, c is equal to one then find a\u00b2 + b\u00b2 + c\u00b2.
\nSolution:
\nEquation of the circle is
\nx2+ y2-2x + 2y +1 = 0
\nPolar of (1, 1) w.r.to the circle is
\nx.1 + y.1 -(x + 1) + (y + 1) + 1 = 0
\nx + y – x – 1 + y + 1 + 1 = 0
\n2y + 1 = 0 …………. (1)
\nGiven equation of the line ax + by + c = 0 ……….. (2)
\nComparing (1) and (2)
\n\\(\\frac{a}{0}=\\frac{b}{2}=\\frac{c}{1}\\) = k
\na = 0, b = 2k, c = k
\na\u00b2 + b\u00b2 + c\u00b2 = 0 + 4k\u00b2 + k\u00b2 = 5k\u00b2
\nH.C.F of (a, b, c) = 1 \u21d2 k = 1
\na\u00b2 + b\u00b2 + c2 = 5(1)\u00b2 = 5<\/p>\n

III.<\/span><\/p>\n

Question 1.
\nFind the coordinates of the point of in-tersection of tangents at the points where x + 4y – 14 = 0 meets the circle x\u00b2 + y\u00b2 – 2x + 3y – 5 = 0.
\nSolution:
\nEquation of the given circle is
\nx\u00b2 + y\u00b2 – 2x + 3y – 5 = 0
\npolar of P(x1<\/sub>, y1<\/sub>) is
\n\"Inter
\n2xx1<\/sub> + 2yy1<\/sub> – 2x – 2x1<\/sub> + 3y + 3y1<\/sub> – 10 = 0
\n2(x1<\/sub> – 1)x + (2y1<\/sub> + 3)y – (2x1<\/sub> – 3y1<\/sub> + 10) = 0 ……….. (1)
\nEquation of QR is x + 4y – 14 = 0 ……… (2)
\nComparing (1) and (2)
\n\"Inter
\n\u2234 Co-ordinates of p are (\\(\\frac{109}{76}\\), \\(\\frac{9}{38}\\))<\/p>\n

Question 2.
\nIf the polar of the points on the circle x\u00b2 + y\u00b2 = a\u00b2 with respect to the circle x\u00b2 + y\u00b2 = b\u00b2 touches the circle x\u00b2 + y\u00b2 = c\u00b2 then prove that a,b,c are in Geometrical progression.
\nSolution:
\n\"Inter
\nP(x1<\/sub>, y1<\/sub>) is a point on the circle x\u00b2 + y\u00b2 = a\u00b2
\n\u2234 x\u00b21<\/sub> + y\u00b21<\/sub> = a\u00b2 ……. (1)
\nPolar of P w.r.to the circle x\u00b2+ y\u00b2 = b\u00b2 is xx1<\/sub> + yy1<\/sub> = b\u00b2
\nThere is a tangent to the circle x\u00b2 + y\u00b2 = c\u00b2
\n\"Inter
\nCross – multiplying we get b2 = ac
\n\u2234 a, b, c are in Geometric progression.<\/p>\n

\"Inter<\/p>\n

Question 3.
\nTangents are drawn to the circle x\u00b2 + y\u00b2 = 16 from the point P(3, 5). Find the area of the triangle formed by these tangents and the chord of contact of P.
\nSolution:
\nEquation of the circle is x\u00b2 + y\u00b2 = 16
\n\"Inter
\nPolar of P (3, 5) is 3x + 5y = 16
\nPL = Length of the perpendicular from P
\n\\(\\frac{|9+25-16|}{\\sqrt{9+25}}=\\frac{18}{\\sqrt{34}}\\)
\nCentre of the circle = C (0, 0)
\nP = Length of the perpendicular from c
\n\"Inter<\/p>\n

Question 4.
\nFind the locus of the point whose polars with respect to the circles x\u00b2 + y\u00b2 – 4x – 4y – 8 = 0 and x\u00b2 + y\u00b2 – 2x + 6y – 2 = 0 are mutually perpendicular.
\nSolution:
\nSuppose P(x, y) is any position the locus Equation of the circles are
\nx\u00b2 + y\u00b2 – 4x – 4y – 8 = 0 ………….. (1)
\nx\u00b2 + y\u00b2 – 2x + 6y – 2 = 0 …………. (2)
\nEquation of the polar of P w.r.to circle (1) is
\nxx1<\/sub> + yy1<\/sub> – 2(x + x1<\/sub>) – 2(y + y1<\/sub>) – 8 = 0
\nx(x1<\/sub> – 2) + y(y1<\/sub> – 2) – (2x1<\/sub> +2y1<\/sub> + 8) = 0 ………… (3)
\nPolar of P w.r. to circle (2) is
\nxx1<\/sub> + yy1<\/sub> – 1(x + x1<\/sub>) – 3(y + y1<\/sub>) – 2 = 0
\nxx1<\/sub> + yy1<\/sub> – x – x1<\/sub> + 3y + 3y1<\/sub> – 2 = 0
\nx(x1<\/sub> – 1) + y(y1<\/sub> + 3) – (x1<\/sub> + 3y1<\/sub> + 2) = 0 ………. (4)
\n(3) and (4) are perpendicular 0
\n\u21d2 a1<\/sub>a2<\/sub> + b1<\/sub>b2<\/sub>
\n(x1<\/sub> – 2) (x1<\/sub> – 1) + (y1<\/sub> – 2) (y1<\/sub> + 3) = 0
\nx\u00b21<\/sub> – 3x1<\/sub> + 2 + y\u00b21<\/sub> + y1<\/sub> – 6 = Q
\nx\u00b21<\/sub> + y\u00b21<\/sub> – 3x1<\/sub> + y1<\/sub> – 4 = 0
\nLocus of p(x1<\/sub>, y1<\/sub>) is x\u00b2 + y\u00b2 – 3x+ y – 4 = 0<\/p>\n

\"Inter<\/p>\n

Question 5.
\nFind the locus of the foot of the perpen\u00acdicular drawn from the origin to any chord of the circle S \u2261 x\u00b2 + y\u00b2 + 2gx + 2fy + c = 0 which subtends a right angle at the origin.
\nSolution:
\n\"Inter
\nSuppose L (x1<\/sub>, y1<\/sub>) is the foot of the per-pendicular from the origin on the chord QL
\nSlope a QL = \\(\\frac{y_{1}}{x_{1}}\\)
\nSlope of QR = \\(\\frac{x_{1}}{y_{1}}\\)
\nEquation of QR is y – y1<\/sub> = \\(\\frac{x_{1}}{y_{1}}\\)(x – x1<\/sub>)
\nyy1<\/sub> – y\u00b21<\/sub> = -xx1<\/sub> + x\u00b21<\/sub>
\nxx1<\/sub> + yy1<\/sub> = x\u00b21<\/sub> + y\u00b21<\/sub>
\nor \\(\\frac{xx_{1}+yy_{1}}{x_{1}^{2}+y_{1}^{2}}\\) = 1 …………. (1)
\nEquation of the circle is
\nx\u00b2 + y\u00b2 + 2gx + 2fy + c = 0 …………. (2)
\nHamogenising (2) with the help of (1), combined equation of
\nQQ, OR is x\u00b2 + y\u00b2 + (2gx + 2fy)
\n\"Inter
\nOQ, OR are perpendicular
\nCo – eff. of x\u00b2 + co- eff of y\u00b2 = 0
\n\"Inter
\n2(x\u00b21<\/sub> + y\u00b21<\/sub>) + 2gx1<\/sub> + 2fy1<\/sub> + c =
\nLocus of L (x1<\/sub>, y1<\/sub>)is
\n2(x\u00b2 + y\u00b2) + 2gx + 2fy + c = 0<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Circle Solutions Exercise 1(d) will help students to clear their doubts quickly. Intermediate 2nd Year Maths 2B Circle Solutions Exercise 1(d) I. Question 1. Find the condition that the tangents drawn from (0, 0) to S \u2261 x\u00b2 + y\u00b2 + …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[14],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35508"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=35508"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35508\/revisions"}],"predecessor-version":[{"id":35509,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35508\/revisions\/35509"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35508"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35508"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35508"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}