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Find the equation of the chord of `x^(2)+y^(2)-6x+10y-9=0` which is bisected at (-2,4)

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To find the equation of the chord of the circle given by \(x^2 + y^2 - 6x + 10y - 9 = 0\) that is bisected at the point \((-2, 4)\), we can follow these steps: ### Step 1: Rewrite the Circle's Equation First, we need to rewrite the equation of the circle in standard form. The given equation is: \[ x^2 + y^2 - 6x + 10y - 9 = 0 \] We can rearrange it to group the \(x\) and \(y\) terms: \[ x^2 - 6x + y^2 + 10y = 9 \] ### Step 2: Complete the Square Next, we complete the square for the \(x\) and \(y\) terms. For \(x\): \[ x^2 - 6x \rightarrow (x - 3)^2 - 9 \] For \(y\): \[ y^2 + 10y \rightarrow (y + 5)^2 - 25 \] Substituting these back into the equation gives: \[ (x - 3)^2 - 9 + (y + 5)^2 - 25 = 9 \] Simplifying this, we have: \[ (x - 3)^2 + (y + 5)^2 - 34 = 9 \] \[ (x - 3)^2 + (y + 5)^2 = 43 \] This shows that the circle has center \((3, -5)\) and radius \(\sqrt{43}\). ### Step 3: Find the Slope of the Radius The radius from the center \((3, -5)\) to the midpoint of the chord \((-2, 4)\) can be used to find the slope of the radius. The slope \(m\) is given by: \[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - (-5)}{-2 - 3} = \frac{9}{-5} = -\frac{9}{5} \] ### Step 4: Find the Slope of the Chord The slope of the chord, which is perpendicular to the radius, is the negative reciprocal of the slope of the radius: \[ m' = -\frac{1}{m} = -\frac{1}{-\frac{9}{5}} = \frac{5}{9} \] ### Step 5: Use the Point-Slope Form to Write the Equation of the Chord Using the point-slope form of the equation of a line, we can write the equation of the chord that passes through the midpoint \((-2, 4)\): \[ y - y_1 = m'(x - x_1) \] Substituting the values: \[ y - 4 = \frac{5}{9}(x + 2) \] ### Step 6: Simplify the Equation Now, we simplify the equation: \[ y - 4 = \frac{5}{9}x + \frac{10}{9} \] Multiplying through by 9 to eliminate the fraction: \[ 9y - 36 = 5x + 10 \] Rearranging gives: \[ 5x - 9y + 46 = 0 \] ### Final Equation Thus, the equation of the chord is: \[ 5x - 9y + 46 = 0 \] ---
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Statement -1 : The equation of chord of the circel x ^(2) + y ^(2) - 6x + 10y - 9=0, which is bisected at the point (-2,4) must be x + y - 2=0. Statement -2 : In notations the equation of chord of the circle S =0 bisected at x _(1), y _(1) must be T = S_(1).

STATEMENT-1 : The equation of chord of circle x^(2) + y^(2) - 6x + 10y - 9 = 0 , which is be bisected at (-2, 4) must be x + y = 2. and STATEMENT-2 : The equation of chord with mid-point (x_(1), y_(1)) to the circle x^(2) + y^(2) = r^(2) is xx_(1) + yy_(1) = x_(1)^(2) + y^(2) .

ARIHANT MATHS-CIRCLE -Exercise (Questions Asked In Previous 13 Years Exam)
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  2. A circle is given by x^2 + (y-1) ^2 = 1, another circle C touches it e...

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  3. If the circles x^2+y^2+2a x+c y+a=0 and points Pa n dQ , then find the...

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  4. A circle touches the x-axis and also touches the circle with center (...

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  5. If a circle passes through the point (a, b) and cuts the circlex x^2+y...

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  6. Let ABCD be a square of side length 2 units. C2 is the circle through ...

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  7. ABCD is a square of side length 2 units. C(1) is the circle touching ...

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  8. ABCD is a square of side length 2 units. C(1) is the circle touching ...

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  9. If the lines 3x-4y-7 = 0 and 2x-3y-5=0 are two diameters of a circle o...

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  10. Let C be the circle with centre (0, 0) and radius 3 units. The equatio...

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  11. Tangents are drawn from the point (17, 7) to the circle x^2+y^2=169, S...

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  12. Consider a family of circles which are passing through the point (-...

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  13. A circle C of radius 1 is inscribed in an equilateral triangle PQR. Th...

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  14. A circle C of radius 1 is inscribed in an equilateral triangle PQR. Th...

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  15. A circle C of radius 1 is inscribed in an equilateral triangle PQR. Th...

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  16. Consider: L1:2x+3y+p-3=0 L2:2x+3y+p+3=0 where p is a real number and...

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  17. The point diametrically opposite to the point P(1, 0) on the circle x^...

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