- AT WHAT POINT IN THE FIRST QUADRANT DOES A LINE WITH AN EQUATION Y = 2X + 1 INTERSECT THE CIRCLE WITH RADIUS 4 AND CENTER (0, 4) AND WE'RE ASKED TO ROUND TO 3 DECIMALS PLACES. SO GRAPHICALLY, WE'RE LOOKING FOR THIS POINT OF INTERSECTION HERE IN THE FIRST QUADRANT. WE'LL TAKE A LOOK AT THE LINE FIRST. NOTICE HOW IT DOES HAVE A Y INTERCEPT OF 1 AND FROM HERE IF WE GO UP 2 UNITS AND RIGHT 1 UNIT WE'RE STILL ON THE SAME LINE VERIFYING THE SLOPE IS 2. AND LOOKING AT THE CIRCLE NOTICE HOW IT HAS A CENTER OF (0, 4) AND THE RADIUS IS 4 UNITS. SO WE'RE GOING TO BEGIN BY DETERMINING THE EQUATION OF THE CIRCLE AND THEN SOLVE THIS AS A SYSTEM OF EQUATIONS. SO USING THIS EQUATION OF A CIRCLE BECAUSE THE RADIUS IS 4 AND THE CENTER IS (0, 4) H IS 0 AND K IS 4. WHICH MEANS THE EQUATION OF THE CIRCLE WOULD BE THE QUANTITY X - 0 SQUARED OR JUST X SQUARED PLUS AGAIN, K IS 4 SO WE HAVE THE QUANTITY Y - 4 SQUARED = R SQUARED WHICH WOULD BE 4 SQUARED OR 16. NOW WE KNOW THE LINEAR EQUATION IS Y = 2X + 1. SO AGAIN, WE'RE GOING TO SOLVE THIS NOW AS A SYSTEM OF EQUATIONS AND WE'RE GOING TO USE SUBSTITUTION. NOTICE HERE Y = 2X + 1 SO WE'LL MAKE THIS SUBSTITUTION FOR Y IN THE FIRST EQUATION. WE'LL SUBSTITUTE 2X + 1 HERE FOR Y. SO THIS WOULD GIVE US THE EQUATION X SQUARED + THE QUANTITY 2X + 1 - 4 SQUARED = 16. LET'S GO AHEAD AND SIMPLIFY THIS. WE'D HAVE X SQUARED + THE QUANTITY 2X THIS WOULD BE 1 - 4 IS - 3 SQUARED = 16. NOTICE NOW WE HAVE AN EQUATION WITH ONE VARIABLE. SO NOW WE'LL EXPAND THIS, COMBINE LIKE TERMS, AND SOLVE THE EQUATION. SO WE'LL HAVE X SQUARED + THE QUANTITY 2X - 3 x THE QUANTITY OF 2X - 3 = 16. HERE WE'LL HAVE FOUR PRODUCTS 1, 2, 3, AND 4. SO WE'LL HAVE X SQUARED AND THEN + 4X SQUARED. THE NEXT PRODUCT WOULD BE - 6X, THEN WE HAVE ANOTHER - 6X THAT'S - 12X AND THEN + 9. X SQUARED + 4X SQUARED WOULD BE 5X SQUARED AND THEN WE HAVE - 12X + 9 = 16. WE HAVE A QUADRATIC EQUATION SO WE'RE GOING TO SET IT EQUAL TO ZERO AND SEE IF IT FACTORS. SO WE'LL SUBTRACT 16 ON BOTH SIDES. THAT WOULD GIVE US 5X SQUARED - 12X AND IF WE SUBTRACT 16 ON BOTH SIDES THAT'S GOING TO GIVE US - 7 = 0. UNFORTUNATELY, THIS IS NOT FACTORABLE. SO NOW WE'LL HAVE TO SOLVE THIS USING THE QUADRATIC FORMULA AND KEEP IN MIND WE'RE LOOKING FOR THE POINT IN THE FIRST QUADRANT SO WE'RE ONLY CONCERNED ABOUT THE POSITIVE SOLUTION TO THIS EQUATION. REMEMBER WHEN USING THE QUADRATIC FORMULA WE'LL HAVE "A" = 5, B = -12, AND C = -7. KEEP IN MIND WE ARE ASKED TO ROUND THE COORDINATES OF THE POINT OF INTERSECTION TO 3 DECIMAL PLACES. LET'S GO AHEAD AND SOLVE THIS ON THE NEXT SLIDE. HERE'S THE QUADRATIC FORMULA IF WE NEED TO REVIEW. WE WOULD HAVE X = -12 + OR - THE SQUARE ROOT OF B SQUARED THAT'S -12 SQUARED - 4 x "A" WHICH IS 5 x C WHICH IS -7. WE'LL DIVIDE ALL THIS BY 2 x "A" OR 2 x 5. SO WE HAVE X = 12 + OR - THE SQUARE ROOT OF-- THIS WOULD BE 144 + 140 THAT'S 284 DIVIDED BY 10. NOW WE DON'T NEED TO SIMPLIFY THIS BECAUSE WE'RE LOOKING FOR A DECIMAL VALUE ROUNDED TO 3 DECIMAL PLACES. BUT WE ARE LOOKING FOR THE POSITIVE VALUE OF X IN THE FIRST QUADRANT WHICH MEANS WE'RE LOOKING FOR THE DECIMAL APPROXIMATION FOR X = 12 + THE SQUARE ROOT OF 284 DIVIDED BY 10. BUT JUST IN CASE YOU WERE CURIOUS, 284 = 4 x 71 SO THAT WOULD BE 12 + OR - THE SQUARE ROOT OF 4 x 71 DIVIDED BY 10 OR 12 + OR - 2 SQUARE ROOT 71 DIVIDED BY 10 AND THEN TO SIMPLIFY THIS WE COULD BREAK IT UP INTO TWO SEPARATE FRACTIONS. 12/10 + OR - 2 SQUARE ROOT 71 DIVIDED BY 10 WHICH WOULD GIVE US 6/5 + OR - SQUARE ROOT 71 DIVIDED BY 5. AND OF COURSE WE COULD WRITE THIS AS A SINGLE FRACTION IF WE WANTED TO. BUT AGAIN, FOR OUR PURPOSES WE'RE ONLY CONCERNED ABOUT THE DECIMAL APPROXIMATION OF THIS 3 DECIMAL PLACES. SO USING A CALCULATOR SO WE'D HAVE OPEN PARENTHESIS FOR THE NUMERATOR 12 + THE SQUARE ROOT 284, CLOSED PARENTHESIS FOR THE SQUARE ROOT, ANOTHER ONE FOR THE NUMERATOR, AND THEN DIVIDED BY 10. SO X IS GOING TO BE APPROXIMATELY 2.885. NOTICE HOW THE 4th DECIMAL PLACE IS A 2 INDICATING WE ROUND DOWN. SO X IS APPROXIMATELY 2.885. WE STILL HAVE TO FIND THE Y COORDINATE THOUGH. WE KNOW THAT THE POINT OF INTERSECTION IN THE FIRST QUADRANT HAS AN X COORDINATE THAT'S APPROXIMATELY 2.885. AND NOW TO FIND THE Y COORDINATE WE'LL PERFORM BACK SUBSTITUTION. LET'S GO AHEAD AND USE THE LINEAR EQUATION WHICH WAS Y = 2X + 1 TO FIND THE Y COORDINATE. SO WE'D HAVE Y = 2 x 2.885 + 1 WHICH COMES OUT TO Y = 6.770. SO THIS WOULD BE THE Y COORDINATE OF THE POINT OF INTERSECTION. AND AGAIN, KEEP IN MIND THESE HAVE BEEN ROUNDED TO 3 DECIMAL PLACES. SO GOING BACK TO OUR GRAPH THIS DOES SEEM TO VERIFY WHAT WE SEE GRAPHICALLY. THE X COORDINATE DOES APPEAR TO BE A LITTLE BIT LESS THAN 3 AND THE Y COORDINATE DOES APPEAR TO BE LESS THAN 7. I HOPE YOU FOUND THIS EXPLANATION HELPFUL.