foreign find the equation of the circle whose Center is minus one comma two because given Center equals to see that equals to minus 1 comma 2 and let P equals to Phi comma 6 radius equals 2 r that equals to cpma distance between C and P is nothing but radius already distance [Music] root of extra minus X1 whole square plus Y2 minus y1 whole Square owner substitution Phi minus of minus plus 1 whole square plus Y 2 minus y 1 whole Square should any five plus one six square six little four square six Exit 36 plus 4 4 16. 12 4 52 therefore equation of circle is Center tells you radius tells you equation of a circle x minus H whole square plus y minus K whole square equals to R square center is x minus of minus plus 1 whole square plus y minus K and the 2 whole square equals to root 52 whole square and then a plus b whole Square x square plus 1 plus 2X plus y square plus 2 square 4 minus 2 2 is a 4y equals to square root cancel 52. x square plus y square plus 2X minus 4 4y it plus 52 itself minus 52 plus 5 12 5 minus 27 times 7 47 but the number symbol minus so M and the minus 47 equals to 0 and when a problem if this is if the circle has the center at 2 comma 3 then find a b and the radius of the circle given equation of a circle given equation of a Circle x square plus y square plus ax plus b y minus 12 equals to 0. radius x square plus y square plus 2 J X plus two f y plus C equals to 0. foreign [Music] formula [Music] root of G square plus F Square minus C root of G Square minus 2 whole Square F Square minus 3 whole Square minus of C minus 12. minus into minus plus 12 9 plus 4 13 13 plus 12 and the other day 25 root 25 and the 5 Square color is square root to cancel that equals to 5 and 3. okay equation of a circle General equation x square plus y square plus 2 J X plus two a five plus C equals to 0 though Circle enter the x square plus y Square minus 4X plus 6y plus a equals to 0 then the general form to conversion x square plus y square plus 2 J X plus 2fy plus C equals to 0 and then T 2 J equals to minus 4 and a J equals to remove 2f equals to 6 and f equals to 3 is equals T into a is equals to 4 enter given radius equals to 4 radius and T root of G square plus F Square minus C equals to 4. g f c value is on both sides foreign [Music] [Music] foreign a equals to 2 comma 3 okay given Circle x square plus y Square minus 8x minus 8y plus 27 equals to 0. x square plus y square plus 2 J X plus two f i plus 3 equals to zero now 2f equals to minus 8 is a f equals to NT to force G value is also F value is also therefore Center equals to minus Z comma minus F that equals to M of Sigma 4 comma 4 minus J minus F color minus into minus plus hypo3 center point also let let b equals to x 2 comma Y2 NTS midpoint formula is C equals to midpoint of a b is the center equals to Center is the midpoint of a because [Music] y1 plus y two y two now in X2 X 1 y 1 and curly substitution by two 3 plus Y2 by 2 . a values x 1 y sorry y 1 x 1 y 1 substitutions to 8 minus 2 x 2 equals to 6. 3 plus Y2 by 2 equals to 4 3 plus y two equals to four to Z 8 Y 2 equals to 8 minus 3 and Y 2 equals 10 the 5 therefore the other end of the diameter is B that equals to x 2 comma Y2 that equals to 6 comma 5 and problems foreign [Music] find the equation of the circle passing through the origin and having the center at minus 4 comma minus 3 x is one yellow first Roman low second problem next find the equation of the circle passing through 2 comma minus 1 having the center at 2 comma 3 x is one yellow first Roman low third problem find the equation of the circle passing through three comma 4 having the center at minus three comma four x is one yellow first terminal fifth problem next find the value of a if 2 x square plus a y Square minus three X plus two y minus 1 equals to 0 represents a circle and also find its radius X is one yellow first Roman low sixth problem next if x square plus y Square minus 4X plus 6y plus six is C equals to 0 represents a circle with radius 6 then find the value of c x is one yellow first Roman load tenth problem next find the center and radius of the circle whose equation is root of 1 plus M square into x square plus y Square minus 2 c x minus 2 m c y equals to zero X is one yellow first sorry seventh one more next find the equation of the circle whose endpoints of a diameter or 4 comma 2 and 1 comma 5 x has one yellow first from a low 12 low fourth one more okay now next obtain the parametric equation of the circle x square plus y square equals to four x has one yellow first Roman low 13th the first one up automatic equation of the circle 2 x square plus 2y square equals to 7 x is one yellow first Roman low 13th low third one next obtain the parametric equation of the circle x minus 3 whole square plus y minus 4 whole square equals to 8 square X is 1 L of first one next applying the parametric equation of the circle x square plus y Square minus six x plus 4y minus 12 equals to zero X is one yellow first Roman low 13th row sixth one next find the equation of the circle which is concentric with x square plus y Square minus 6 x minus 4y minus 12 equals to 0 and passing through minus 2 comma 14 x is one yellow second normal fifth problem okay now next find the power of the point p with respect to the Circle S equals to 0 when P equals to minus 1 comma 1 and S equals to x square plus Y is Square minus 6X plus 4y minus 1 x i is 1 below first Roman low second low second one okay now next find the length of the tangent from P to the Circle S equals to zero when P equals to minus 2 comma 5 and S equals to x square plus y Square minus 25 x is 1 below first Roman look third low first one okay now next if the length of the tangent from Phi comma 4 to the circle x square plus y square plus 2 k y equals to 0 is 1 then find k x is 1 below second Roman low first one next if the length of the tangent from 2 comma fine to the circle x square plus y Square minus 5x plus 4y plus k equals to 0 is root 37 then find k x is 1 below second Roman low second problem okay now next 21st one question shouldn't be find the equation of normal at P of the Circle S equals to 0 where P equals to 3 comma 5 and is equals to an A okay equation equation of a normal concurring given s equals to x square plus y Square minus 10x minus 2 y plus 6 is x square plus y square plus 2 J X Plus 2fy plus C equals to 0 of the circle is equals to 0 so equals to zero beta so 2G equals minus 10 and a J equals 10 that to Phi Za G value is given point therefore equation of normal is important x minus X1 into y1 plus f minus y minus y1 into X1 plus g equals to 0 okay now equation of a normal formula NT x minus x 1 into y1 plus F minus y minus y1 into X1 plus g equals to 0 is X1 y1 is x minus X1 and then the three foreign 4X minus 12 plus 2y minus 10 equals to 0. 4X plus 2y 10 12 and that will be 22 minus 22 equals to zero two common nature two common they need to do that equals to 0 therefore equation of a circle is 2 multiplier than it said because the divided 0 anything 0 so M and E 2 X Plus y minus 11 equals to 0 and the equation of a normal life equation of a circle x square plus y square plus 2 J X plus two f y plus C equals to 0 though x minus x 1 into y one plus F minus of Y minus y 1 into x 1 plus g equals to zero if x 1 y 1 is is to a square on the line X cos Alpha plus y sine Alpha equals to P length of the current Formula G to root R square minus d square Max problems R square minus D Square D and 10 perpendicular distance from the uh Center okay now and as always given Circle enter the x square plus y square equals to a square if a model on the x square plus y square equals to R square model on the until origin first exercise given line enter the excuse Alpha plus y sine Alpha e plus b type question minus P equals to 0. LX Plus m y plus n equals to 0 okay sine Alpha perpendicular distance from Center and then the modulus of Emma today LX Plus m y plus n by root of L square plus M square formula length R square minus equation of a circle is n equals to zero models to cos Alpha m equals to sine alpha n equals to minus b d length perpendicular distance from the center and take the x is modulus of LX Plus m y plus n by root of L square plus M Square perpendicular distance from the Centric of pharmace modulus of LX Plus m y plus n by root of L square plus M Square substitution cos Alpha into X Y value is so into 0 plus M1 20 sine alpha y n t zero nnt minus P by root of L Square cos Square Alpha plus M Square sine Square Alpha therefore length of the Curve equals to 2 root R square minus d square that equals to 2 root a so a square minus D and tenth of P perpendicular distance from the center Farm lion T modulus of LX Plus m y plus n by root of L square plus M Square equal to x value y Value Center points means substitutions [Music] thank you